JEE Magnetism & Magnetic Materials Questions

Velocity has components:

parallel to B (along k):
v∥ = 2 × 10⁻² m/s

perpendicular to B (in i direction):
v⊥ = 3 × 10⁻² m/s

motion is helical

time period:

$$T=\frac{2\pi m}{qB}$$

given:

m = 5 mg = 5 × 10⁻⁶ kg
q = 5π × 10⁻⁶ C
B = 0.1

$$T=\frac{2\pi\times5\times10^{-6}}{5\pi\times10^{-6}\times0.1}=\frac{2}{0.1}=20\text{ s}$$

time for 5 revolutions:

$$t=5T=100\text{ s}$$

distance along k:

$$a=v∥\times t=2\times10^{-2}\times100=2$$

The de-Broglie wavelength of a particle is defined as
$$\lambda = \frac{h}{p} = \frac{h}{m\,v}$$
where $$v$$ is the magnitude (speed) of the velocity vector.

At $$t = 0$$ the electron has velocity $$\vec v_0 = v_0\hat i$$ and de-Broglie wavelength $$\lambda_0 = \dfrac{h}{m\,v_0}$$.

After entering the magnetic field $$\vec B = B_0\hat j$$ the electron experiences the Lorentz force
$$\vec F = q\,\vec v \times \vec B$$
For an electron, $$q = -e$$ (magnitude $$e$$). At any instant the velocity has an $$\hat i$$ component only, so
$$\vec F = (-e)\,(v_x\hat i)\times (B_0\hat j) = -e\,v_x B_0\,(\hat i \times \hat j) = -e\,v_x B_0\,\hat k$$
The force is always perpendicular to the instantaneous velocity.

Since $$\vec F \perp \vec v$$, the magnetic force does no work:
$$\text{Power} = \vec F\cdot\vec v = 0$$

Hence the kinetic energy $$\frac12 m v^2$$ remains constant, so the speed $$v$$ of the electron stays equal to its initial value $$v_0$$ for all time.

Because the speed (and therefore the magnitude of momentum $$p = m v$$) is constant, the de-Broglie wavelength also stays constant:
$$\lambda = \frac{h}{m v} = \frac{h}{m v_0} = \lambda_0$$

Thus, after any time $$t$$, the wavelength is still $$\lambda_0$$.

Final answer: $$\lambda_0$$   →   Option D

When an electron is projected perpendicular to a uniform magnetic field $$B$$, it moves in a circular path due to the magnetic force acting as the centripetal force. The force balance equation is given by:

$$ e v B = \frac{m v^2}{r} $$

where $$e$$ is the charge of the electron, $$v$$ is its speed, $$m$$ is its mass, and $$r$$ is the radius of the circular path. Simplifying this equation:

$$ e B = \frac{m v}{r} $$

Rearranging, we get:

$$ m v = e B r \quad \text{(1)} $$

Bohr's quantization condition states that the angular momentum is quantized and given by:

$$ m v r = n \frac{h}{2\pi} \quad \text{(2)} $$

where $$n$$ is the principal quantum number and $$h$$ is Planck's constant. The first excited state corresponds to $$n = 2$$ (since the ground state is $$n = 1$$). Substituting $$n = 2$$ into equation (2):

$$ m v r = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \quad \text{(3)} $$

We now have two equations: equation (1) $$m v = e B r$$ and equation (3) $$m v r = \frac{h}{\pi}$$. Substituting equation (1) into equation (3):

$$ (e B r) \times r = \frac{h}{\pi} $$

Simplifying:

$$ e B r^2 = \frac{h}{\pi} $$

Solving for $$r^2$$:

$$ r^2 = \frac{h}{\pi e B} $$

Taking the square root of both sides:

$$ r = \sqrt{\frac{h}{\pi e B}} $$

This is the radius of the electronic orbit in the first excited state. Comparing with the given options, this expression matches option A, $$\sqrt{\frac{h}{\pi eeB}}$$, when interpreted as $$\sqrt{\frac{h}{\pi e B}}$$ (considering "ee" as a typographical error for "e").

Thus, the correct option is A.

When a charged particle of charge $$q$$ and linear momentum $$p$$ enters a uniform magnetic field $$\mathbf{B}$$ perpendicular to the field, the magnetic force $$|q|vB$$ provides the required centripetal force $$\dfrac{mv^{2}}{r}$$.

Substituting $$p = mv$$ and $$v = \dfrac{p}{m}$$, the radius (called radius of curvature) is obtained as
$$r = \dfrac{p}{|q|B}$$ $$-(1)$$

Case 1: Oxide ion $$O^{2-}$$ versus proton $$H^{+}$$
For $$O^{2-}: |q| = 2e \quad(\text{two electronic charges})$$
For $$H^{+}: |q| = e$$
Both ions are given to have the same momentum $$p$$ and the same field $$B$$, so using $$(1)$$:

$$r_{O^{2-}} = \dfrac{p}{2eB}, \qquad r_{H^{+}} = \dfrac{p}{eB} = 2\,r_{O^{2-}}$$

Thus $$r_{O^{2-}} \lt r_{H^{+}}$$, i.e. the oxide ion describes a circle of smaller radius (larger bending) than the proton. Consequently its curvature radius is smaller. Hence Assertion A is correct.

Case 2: Proton versus electron with the same linear momentum
For a proton $$|q_p| = e$$ and for an electron $$|q_e| = e$$. With equal momenta $$p$$, equation $$(1)$$ gives

$$r_p = \dfrac{p}{eB}, \qquad r_e = \dfrac{p}{eB}$$

The radii are identical; the proton does not form a path of smaller radius than the electron. Therefore Reason R is false.

Conclusion: Assertion A is true, Reason R is false. The correct choice is Option A.

The magnetic field at the centre of a square loop carrying current can be found by applying the Biot-Savart law to one side and then using symmetry. Given that I = 5 A, a = 1/√2 m and μ₀ = 4π×10⁻⁷ T m A⁻¹, consider a single straight segment of the loop. For a finite straight wire the field at a perpendicular distance d, subtending angles θ₁ and θ₂ at the observation point, is $$B_{\text{wire}} = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2).$$ In our square loop the distance from the centre to each side is d = a/2, and each side subtends equal angles θ₁ = θ₂ = 45° at the centre. Substituting these values gives $$B_1 = \frac{\mu_0 I}{4\pi\,(a/2)}\bigl(\sin45°+\sin45°\bigr) = \frac{\mu_0 I}{2\pi a}\times\frac{2}{\sqrt2} = \frac{\sqrt2\,\mu_0 I}{2\pi a}.$$ By symmetry all four sides contribute equally and their fields add in the same direction, so the total field is $$B = 4B_1 = 4\;\frac{\sqrt2\,\mu_0 I}{2\pi a} = \frac{2\sqrt2\,\mu_0 I}{\pi a}.$$ Substituting I = 5 A, a = 1/√2 m and μ₀ = 4π×10⁻⁷ T m A⁻¹ yields $$B = \frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi\times(1/\sqrt2)}.$$ The π factors cancel, giving $$B = \frac{2\sqrt2\times4\times5\times10^{-7}}{1/\sqrt2} = 40\sqrt2\times10^{-7}\times\sqrt2 = 40\times2\times10^{-7} = 8\times10^{-6}\text{ T}.$$ Therefore, the magnetic field at the centre of the square loop is $$B = 8\times10^{-6}\text{ T},$$ and the required value of p is 8.

When a charged particle is fired perpendicular to a uniform magnetic field, it traces a circular path. The time required for the particle to return to its original location for the first time is exactly the time period of this circular motion.

The magnetic force provides the centripetal force: $$qvB = \frac{mv^{2}}{r}$$

This gives the radius $$r = \frac{mv}{qB}$$.

The time period is $$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$$

The time period does not depend on the speed of the particle.

Substituting into the time period formula:

$$T = \frac{2 \times 3.14 \times 16 \times 10^{-6}}{1.6 \times 10^{-6} \times 6.28}$$

$$T = \frac{100.48 \times 10^{-6}}{10.048 \times 10^{-6}} = \frac{100.48}{10.048} = 10 \text{ s}$$

So, the answer is $$10$$.

We are told a proton moves undeflected in crossed electric and magnetic fields at $$v = 2 \times 10^5$$ m/s. When the electric field is switched off, the proton moves in a circle of radius $$r = 2$$ cm. We need to find the electric field.

Condition for undeflected motion in crossed fields

When a charged particle moves undeflected through crossed electric and magnetic fields, the electric force and magnetic force balance each other:

$$qE = qvB$$

$$E = vB \quad \ldots (1)$$

Find $$B$$ from the circular motion when $$E$$ is switched off

When the electric field is switched off, only the magnetic force acts, providing the centripetal force for circular motion:

$$qvB = \frac{mv^2}{r}$$

$$B = \frac{mv}{qr} \quad \ldots (2)$$

Substituting given values ($$m = 1.6 \times 10^{-27}$$ kg, $$v = 2 \times 10^5$$ m/s, $$q = 1.6 \times 10^{-19}$$ C, $$r = 2$$ cm $$= 0.02$$ m):

$$B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02}$$

$$B = \frac{3.2 \times 10^{-22}}{3.2 \times 10^{-21}} = \frac{1}{10} = 0.1 \text{ T}$$

Calculate the electric field using equation (1)

$$E = vB = 2 \times 10^5 \times 0.1 = 2 \times 10^4 \text{ N/C}$$

Therefore, $$x = 2$$.

The answer is 2.

$$T = \frac{2\pi m_e}{|q_e| \mu_0 n I}$$

$$75 \times 10^{-9} = \frac{(2\pi)(9 \times 10^{-31})}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$$

$$75 \times 10^{-9} = \frac{9 \times 10^{-31}}{1.6 \times 10^{-19} \times 2 \times 10^{-7} \times n \times 1.5}$$

$$75 \times 10^{-9} = \frac{9 \times 10^{-31}}{4.8 \times 10^{-26} \times n} \implies n = 250\text{ m}^{-1}$$

Given: Current $$I = 2$$ A, resistance $$R = 1\Omega$$, resistivity $$\rho = 2 \times 10^{-6} \Omega$$m, area $$A = 10$$ mm$$^2 = 10 \times 10^{-6}$$ m$$^2$$, and mass $$m = 500$$ g = 0.5 kg.

First, we find the length of the wire using the relation $$R = \frac{\rho L}{A}$$ which gives $$L = \frac{RA}{\rho} = \frac{1 \times 10 \times 10^{-6}}{2 \times 10^{-6}} = 5 \text{ m}$$.

Next, the magnetic force must balance the weight of the wire, so $$BIL = mg$$ and hence $$B = \frac{mg}{IL} = \frac{0.5 \times 10}{2 \times 5} = \frac{5}{10} = 0.5 \text{ T}$$.

Therefore, $$B = 0.5$$ T $$= 5 \times 10^{-1}$$ T.

The answer is 5.

A circular coil with 200 turns, area $$2.5 \times 10^{-4}$$ m² carrying $$100 \mu$$A current is in a uniform magnetic field of 1 T. We need the work to rotate it 90° from aligned to perpendicular orientation.

The magnetic dipole moment is $$M = NIA = 200 \times 100 \times 10^{-6} \times 2.5 \times 10^{-4}$$ which gives $$M = 200 \times 10^{-4} \times 2.5 \times 10^{-4} = 5 \times 10^{-6} \text{ A m}^2$$.

The potential energy of a magnetic dipole in a field is $$U = -MB\cos\theta$$. Initially ($$\theta = 0°$$), $$U_i = -MB\cos 0° = -MB$$ and finally ($$\theta = 90°$$), $$U_f = -MB\cos 90° = 0$$.

Therefore the work done is $$W = U_f - U_i = 0 - (-MB) = MB$$ which equals $$W = 5 \times 10^{-6} \times 1 = 5 \times 10^{-6} \text{ J} = 5 \text{ }\mu\text{J}$$.

The answer is $$\boxed{5}$$ $$\mu$$J.

Find the torsional constant $$k = x \times 10^{-5}$$ N-m/rad of a galvanometer.

For a moving coil galvanometer, at equilibrium the magnetic torque equals the restoring torque:

$$ NBIA = k\theta $$

where $$N$$ = number of turns, $$B$$ = magnetic field, $$I$$ = current, $$A$$ = area of each turn, $$k$$ = torsional constant, $$\theta$$ = deflection angle.

$$N = 100$$, $$B = 0.01$$ T, $$I = 10$$ mA $$= 10 \times 10^{-3}$$ A, $$A = 2.0$$ cm$$^2$$ $$= 2.0 \times 10^{-4}$$ m$$^2$$, $$\theta = 0.05$$ rad.

$$ k = \frac{NBIA}{\theta} = \frac{100 \times 0.01 \times 10 \times 10^{-3} \times 2.0 \times 10^{-4}}{0.05} $$

Numerator: $$100 \times 0.01 = 1$$

$$1 \times 10 \times 10^{-3} = 10^{-2}$$

$$10^{-2} \times 2.0 \times 10^{-4} = 2.0 \times 10^{-6}$$

$$ k = \frac{2.0 \times 10^{-6}}{0.05} = \frac{2.0 \times 10^{-6}}{5 \times 10^{-2}} = 0.4 \times 10^{-4} = 4 \times 10^{-5} \text{ N-m/rad} $$

So $$x = 4$$.

The answer is 4.

A regular hexagon is formed by bending a wire of length $$4\pi$$ m. A current $$I = 4\pi\sqrt{3}$$ A flows through it, and we wish to find the magnetic field at the centre.

A regular hexagon has 6 sides, so its side length is $$a = \frac{4\pi}{6} = \frac{2\pi}{3}$$ m. The perpendicular distance from the centre to each side (the apothem) is $$d = \frac{a\sqrt{3}}{2} = \frac{2\pi}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{3}$$ m.

Each side subtends an angle of $$60°$$ at the centre, giving a half-angle of $$30°$$. The magnetic field due to a finite straight conductor at perpendicular distance $$d$$ is given by $$B_1 = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$$, and since $$\alpha_1 = \alpha_2 = 30°$$, this becomes $$B_1 = \frac{\mu_0 I}{4\pi d}(2\sin 30°) = \frac{\mu_0 I}{4\pi d}$$.

The total magnetic field from all six sides is $$B = 6 \times \frac{\mu_0 I}{4\pi d} = \frac{6\mu_0 I}{4\pi d} = \frac{6 \times 4\pi \times 10^{-7} \times 4\pi\sqrt{3}}{4\pi \times \frac{\pi\sqrt{3}}{3}} = \frac{6 \times 4\pi\sqrt{3} \times 10^{-7}}{\frac{\pi\sqrt{3}}{3}} = \frac{6 \times 4\pi\sqrt{3} \times 3}{\pi\sqrt{3}} \times 10^{-7} = 72 \times 10^{-7}$$ T.

The value of $$x$$ is 72.

We need to find the number of turns $$m$$ in a solenoid, given its dimensions and the magnetic field it produces.

The magnetic field inside a long solenoid is given by $$B = \mu_0 n I = \mu_0 \frac{m}{L} I$$ where $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A is the permeability of free space, $$n = m/L$$ is the number of turns per unit length, $$m$$ is the total number of turns, $$L$$ is the length of the solenoid, and $$I$$ is the current.

Substituting the given values $$B = 6.28 \times 10^{-3}$$ T, $$L = 0.5$$ m, and $$I = 5$$ A into this formula gives $$6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times \frac{m}{0.5} \times 5$$.

This simplifies to $$6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times 10m$$ and hence $$6.28 \times 10^{-3} = 4\pi \times 10^{-6} \times m$$. Using $$\pi \approx 3.14$$, we have $$6.28 \times 10^{-3} = 4 \times 3.14 \times 10^{-6} \times m = 12.56 \times 10^{-6} \times m$$. From this, $$m = \frac{6.28 \times 10^{-3}}{12.56 \times 10^{-6}} = \frac{6.28}{12.56} \times 10^{3} = 0.5 \times 10^{3} = 500$$.

The answer is 500.

Let the square loop have its vertical sides located at positions $$x$$ and $$x + L$$, with side length $$L = 2\text{ m}$$.

$$F_{\text{net}} = F_{\text{right}} - F_{\text{left}}$$

$$\implies F_{\text{net}} = I L B_0 [1 + 4(x+L)] - I L B_0 [1 + 4x]$$ $$\implies F_{\text{net}} = 4 I B_0 L^2$$

$$\implies F_{\text{net}} = 4 \times 2 \times 5 \times (2)^2 = 160\text{ N}$$

$$\vec{F} = q(\vec{v} \times \vec{B})$$

$$\vec{v} \times \vec{B} = (3\hat{i} + 5\hat{j}) \times (B_0\hat{i} + 2B_0\hat{j})$$

$$\implies \vec{v} \times \vec{B} = 3(2B_0)\hat{k} + 5B_0(-\hat{k}) = 6B_0\hat{k} - 5B_0\hat{k} = B_0\hat{k}$$

$$\vec{F} = 5e\hat{k}$$

$$5e\hat{k} = -e(B_0\hat{k})$$

$$\implies 5 = -B_0 \implies B_0 = -5\text{ T}$$ $$\implies |B_0| = 5$$

An electron with KE = 5 eV enters a magnetic field of 3 $$\mu$$T. Find the electric field $$E$$ for the electron to move undeflected.

$$KE = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times KE}{m}}$$

Convert KE to Joules: $$5 \text{ eV} = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19}$$ J.

$$v = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{1.778 \times 10^{12}}$$

$$v \approx 1.33 \times 10^6 \text{ m/s}$$

$$qE = qvB \implies E = vB$$

$$E = 1.33 \times 10^6 \times 3 \times 10^{-6} = 3.99 \approx 4 \text{ N/C}$$

The correct answer is 4 NC$$^{-1}$$.

We need to find the current required to demagnetise a magnet placed inside a solenoid, given the coercivity of the magnet.

Coercivity is the intensity of the magnetising field ($$H$$) required to reduce the magnetisation of a material to zero after it has been magnetised. In other words, to demagnetise the magnet, we need to apply a magnetic field intensity equal to its coercivity, which is $$H = 5 \times 10^3$$ A/m.

Since the magnetic field intensity inside a solenoid is given by $$H = nI = \frac{N}{L} \times I$$, where $$N$$ is the total number of turns, $$L$$ is the length of the solenoid, and $$I$$ is the current, we can use this relation to determine the required current.

Next, substituting $$N = 150$$ turns, $$L = 30$$ cm $$= 0.3$$ m, and $$H = 5 \times 10^3$$ A/m into the expression gives $$5 \times 10^3 = \frac{150}{0.3} \times I$$, which simplifies to $$5 \times 10^3 = 500 \times I$$. From this, $$I = \frac{5 \times 10^3}{500} = 10 \text{ A}$$.

The answer is 10 A.

$$\text{Magnetic field contribution from straight wire segments:}$$ $$B_{\text{straight}} = 0$$

$$\text{Total magnetic field at the centre O:}$$ $$B = B_1 + B_2 = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2}$$

$$B = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$

$$B = \frac{(4\pi \times 10^{-7})(4)}{4} \left( \frac{1}{2\pi} + \frac{1}{4\pi} \right)$$

$$B = (4\pi \times 10^{-7}) \left( \frac{3}{4\pi} \right) = 3 \times 10^{-7}\ \text{T}$$

$$\alpha \times 10^{-7} = 3 \times 10^{-7} \implies \alpha = 3$$ 

For horizontal segments parallel to x-axis:

$$d\vec{F} = I(dx\,\hat{i} \times \vec{B})$$

$$\vec{F}_{\text{top}} + \vec{F}_{\text{bottom}} = 0$$

For vertical segment at $$x_1 = 2$$:

$$\vec{F}_1 = I\vec{L}_1 \times \vec{B}(x_1)$$

$$\vec{F}_1 = 0.5(-0.5\hat{j}) \times [0.2(1 + 2(2))\hat{k}]$$

$$\vec{F}_1 = -0.25\hat{j} \times 1\hat{k} = -0.25\hat{i}$$

For vertical segment at $$x_2 = 2 + 0.5 = 2.5$$:

$$\vec{F}_2 = I\vec{L}_2 \times \vec{B}(x_2)$$

$$\vec{F}_2 = 0.5(0.5\hat{j}) \times [0.2(1 + 2(2.5))\hat{k}]$$

$$\vec{F}_2 = 0.25\hat{j} \times 1.2\hat{k} = 0.30\hat{i}$$

$$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2$$

$$\vec{F}_{\text{net}} = -0.25\hat{i} + 0.30\hat{i} = 0.05\hat{i}$$

$$|\vec{F}_{\text{net}}| = 0.05\text{ N} = 50\text{ mN}$$

Magnetic potential on axis: $$V = \frac{\mu_0}{4\pi}\frac{M\cos\theta}{r^2}$$. On axis, $$\theta = 0$$:

$$V = \frac{\mu_0 M}{4\pi r^2} = 10^{-7} \times \frac{M}{(0.2)^2} = 10^{-7} \times 25M$$

$$1.5 \times 10^{-5} = 25 \times 10^{-7} M$$. $$M = \frac{1.5 \times 10^{-5}}{25 \times 10^{-7}} = 6$$ Am².

The answer is $$\boxed{6}$$.

A circular coil of $$N$$ turns and radius $$R$$ produces, at its centre, a magnetic field along the axis of the coil given by
$$B = \frac{\mu_0 N I}{2R} \quad -(1)$$
where $$I$$ is the current through the coil.

For coil $$P$$:
Number of turns, $$N_P = 100$$
Current, $$I_P = 1 \, \text{A}$$
Radius, $$R = \pi \, \text{cm} = \pi \times 10^{-2} \, \text{m}$$

Substituting in $$(1)$$,
$$B_P = \frac{\mu_0 N_P I_P}{2R}$$
$$= \frac{(4\pi \times 10^{-7}) (100)(1)}{2(\pi \times 10^{-2})}$$
$$= \frac{4\pi \times 10^{-5}}{2\pi \times 10^{-2}}$$
$$= 2 \times 10^{-3} \, \text{T}$$
$$= 2 \, \text{mT} \quad -(2)$$

For coil $$Q$$:
Number of turns, $$N_Q = 100$$ (same as $$P$$)
Current, $$I_Q = 2 \, \text{A}$$

Again using $$(1)$$,
$$B_Q = \frac{\mu_0 N_Q I_Q}{2R}$$
$$= \frac{(4\pi \times 10^{-7}) (100)(2)}{2(\pi \times 10^{-2})}$$
$$= \frac{8\pi \times 10^{-5}}{2\pi \times 10^{-2}}$$
$$= 4 \times 10^{-3} \, \text{T}$$
$$= 4 \, \text{mT} \quad -(3)$$

The planes of the two coils are mutually perpendicular, so the fields $$B_P$$ and $$B_Q$$ are perpendicular to each other. Hence, the resultant field at the common centre is obtained by vector addition using the Pythagoras theorem:

$$B_{\text{res}} = \sqrt{B_P^{\,2} + B_Q^{\,2}}$$
$$= \sqrt{(2 \, \text{mT})^{2} + (4 \, \text{mT})^{2}}$$
$$= \sqrt{4 + 16} \, \text{mT}$$
$$= \sqrt{20} \, \text{mT}$$

The question states $$B_{\text{res}} = \sqrt{x} \, \text{mT}$$. Comparing, $$x = 20$$.

Hence, the required value is 20.

$$B_{\text{net}} = B_1 + B_2 = 2 \times \left(\frac{\mu_0 I}{2\pi r}\right)$$

$$I = 10\text{ A},\quad d = 5.0\text{ cm} \implies r = \frac{d}{2} = 2.5\text{ cm} = 2.5 \times 10^{-2}\text{ m}$$

$$B_{\text{net}} = 2 \times \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 2.5 \times 10^{-2}} \implies B_{\text{net}} = 1.6 \times 10^{-4}\text{ T} = 160\ \mu\text{T}$$

$$B_A = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 (2I)}{2\pi (3r)} = \frac{\mu_0 I}{2\pi r} \left(1 + \frac{2}{3}\right) = \frac{5\mu_0 I}{6\pi r}$$

$$B_C = \frac{\mu_0 I}{2\pi (3r)} + \frac{\mu_0 (2I)}{2\pi r} = \frac{\mu_0 I}{2\pi r} \left(\frac{1}{3} + 2\right) = \frac{7\mu_0 I}{6\pi r}$$

$$\frac{B_A}{B_C} = \frac{5}{7} \implies x = 5$$

A coil of 200 turns and area $$0.20 \text{ m}^2$$ is rotated at half a revolution per second in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation. We need to find $$\beta$$ where the maximum voltage is $$\frac{2\pi}{\beta}$$ V.

The EMF induced in a rotating coil in a magnetic field is:

$$\varepsilon = NAB\omega\sin(\omega t)$$

The maximum EMF is:

$$\varepsilon_{\max} = NAB\omega$$

The coil rotates at $$\frac{1}{2}$$ revolution per second, so:

$$\omega = 2\pi \times \frac{1}{2} = \pi \text{ rad/s}$$

$$\varepsilon_{\max} = 200 \times 0.20 \times 0.01 \times \pi$$

$$= 200 \times 0.002 \times \pi$$

$$= 0.4\pi$$

$$= \frac{2\pi}{5}$$

Comparing $$\frac{2\pi}{5}$$ with $$\frac{2\pi}{\beta}$$:

$$\beta = 5$$

The answer is 5.

We need to find the magnetic moment of a straight magnetic strip after it is bent into a semicircular shape.

Original magnetic moment: $$M = 44 \text{ Am}^2$$

$$\pi = \frac{22}{7}$$

For a bar magnet (or magnetic strip), the magnetic moment is defined as:

$$M = m \times L$$

where $$m$$ is the pole strength and $$L$$ is the length of the strip (distance between the poles). Here:

$$44 = m \times L \implies m = \frac{44}{L}$$

When the strip of length $$L$$ is bent into a semicircular shape, the length of the strip equals the semicircular arc:

$$\pi R = L \implies R = \frac{L}{\pi}$$

where $$R$$ is the radius of the semicircle.

In a semicircle, the two ends of the strip (which carry the magnetic poles) are at diametrically opposite points. The distance between them is the diameter:

$$d = 2R = \frac{2L}{\pi}$$

The pole strength $$m$$ remains unchanged (it is an intrinsic property of the magnet). The new magnetic moment is:

$$M' = m \times d = m \times \frac{2L}{\pi} = \frac{2mL}{\pi} = \frac{2M}{\pi}$$

$$M' = \frac{2 \times 44}{\pi} = \frac{88}{\frac{22}{7}} = \frac{88 \times 7}{22} = \frac{616}{22} = 28 \text{ Am}^2$$

The answer is 28 Am$$^2$$.

Magnetic field at center of a square loop of side $$a$$ with current $$I$$:

Each side contributes $$B_1 = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$$ where $$d = a/2$$ and $$\alpha_1 = \alpha_2 = 45°$$.

$$B_1 = \frac{\mu_0 I}{4\pi (a/2)} \times 2\sin 45° = \frac{\mu_0 I}{2\pi a} \times \sqrt{2} = \frac{\sqrt{2}\mu_0 I}{2\pi a}$$

Total for 4 sides: $$B = 4B_1 = \frac{4\sqrt{2}\mu_0 I}{2\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}$$

With $$I = 5$$A, $$a = 1$$m:

$$B = \frac{2\sqrt{2} \times 4\pi \times 10^{-7} \times 5}{\pi \times 1} = 2\sqrt{2} \times 4 \times 5 \times 10^{-7} = 40\sqrt{2} \times 10^{-7}$$

So $$X = 40$$.

Therefore, the answer is $$\boxed{40}$$.

The force per unit length on a current-carrying conductor in a magnetic field is given by $$\frac{F}{L} = I \times B \times \sin\theta$$, where $$\theta$$ is the angle between the direction of current and the magnetic field. The horizontal component of Earth’s magnetic field is $$B_H = 3.5 \times 10^{-5} \text{ T}$$ directed towards geographic North, and the current is $$I = \sqrt{2} \text{ A}$$ flowing from South-East to North-West. The angle between the North-West direction and the North direction is $$45°$$, so the angle between the current direction (SE to NW) and $$B_H$$ (pointing North) is $$\theta = 45°$$.

Substituting into the formula gives

$$\frac{F}{L} = I \times B_H \times \sin 45°$$

$$\frac{F}{L} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}}$$

$$\frac{F}{L} = 3.5 \times 10^{-5} = 35 \times 10^{-6} \text{ N/m}$$

Therefore, the correct answer is $$35$$.

We are given that $$\lambda = 5000$$ Å $$= 5 \times 10^{-7}$$ m, $$d = 0.3$$ mm $$= 3 \times 10^{-4}$$ m, and $$D = 200$$ cm $$= 2$$ m.

The position of the $$n$$-th maximum in YDSE is given by

$$x_n = \frac{n\lambda D}{d}$$

For the third maximum ($$n = 3$$), we then have

$$x_3 = \frac{3 \times 5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = \frac{30 \times 10^{-7}}{3 \times 10^{-4}} = 10 \times 10^{-3} = 10 \text{ mm}$$

Hence, the third maximum is located at 10 mm from the central maximum.

Current sensitivity of a galvanometer is given by:

$$ I_s = \frac{nBA}{c} $$

where $$n$$ = number of turns, $$B$$ = magnetic field, $$A$$ = area of coil, $$c$$ = restoring torque per unit twist.

To increase current sensitivity by 50%, we increase $$n$$ by 50%.

Voltage sensitivity is given by:

$$ V_s = \frac{I_s}{R} = \frac{nBA}{cR} $$

When the number of turns increases by 50%, the resistance of the coil also increases by 50% (since resistance is proportional to the length of wire, which is proportional to $$n$$).

New voltage sensitivity:

$$ V_s' = \frac{1.5n \times BA}{c \times 1.5R} = \frac{nBA}{cR} = V_s $$

The voltage sensitivity remains unchanged. The percentage change is 0%.

A long conducting wire of total length $$L$$ carrying current $$I$$ is bent into a circular coil of $$N$$ turns, and then into a coil of $$n$$ turns. We need the ratio of magnetic fields at the centres.

$$ B = \frac{\mu_0 N I}{2R} $$

where $$N$$ is the number of turns and $$R$$ is the radius.

For $$N$$ turns: $$L = N \times 2\pi R_1$$, so $$R_1 = \frac{L}{2\pi N}$$

For $$n$$ turns: $$L = n \times 2\pi R_2$$, so $$R_2 = \frac{L}{2\pi n}$$

$$ B_1 = \frac{\mu_0 N I}{2R_1} = \frac{\mu_0 N I}{2} \times \frac{2\pi N}{L} = \frac{\mu_0 \pi N^2 I}{L} $$

$$ B_2 = \frac{\mu_0 n I}{2R_2} = \frac{\mu_0 n I}{2} \times \frac{2\pi n}{L} = \frac{\mu_0 \pi n^2 I}{L} $$

$$ \frac{B_1}{B_2} = \frac{N^2}{n^2} $$

The ratio of magnetic field in the first case to the second case is Option C: $$N^2 : n^2$$.

The magnetic field inside a solenoid is given by:

$$B = \mu_0 n I$$

where $$n$$ is the number of turns per unit length and $$I$$ is the current.

Given: $$n = 70$$ turns/cm $$= 7000$$ turns/m, $$I = 2.0$$ A, $$\mu_0 = 4\pi \times 10^{-7}$$ T m A$$^{-1}$$.

$$B = 4\pi \times 10^{-7} \times 7000 \times 2.0$$

$$= 4\pi \times 14000 \times 10^{-7}$$

$$= 56000\pi \times 10^{-7}$$

$$= 56\pi \times 10^{-4}$$

$$= 56 \times 3.14159 \times 10^{-4}$$

$$\approx 175.93 \times 10^{-4}$$

$$\approx 176 \times 10^{-4}$$ T

The correct answer is Option 2: $$176 \times 10^{-4}$$ T.

$$F_{\text{gravity}} = mg = (1 \times 10^{-3}\text{ kg}) \times 10\text{ m/s}^2 = 10^{-2}\text{ N}$$

$$F_{\text{magnetic}} = \left(\frac{V}{R}\right) \cdot l \cdot B$$

$$\frac{V \cdot l \cdot B}{R} = mg$$

$$V = \frac{mg \cdot R}{l \cdot B}$$

$$V = \frac{10^{-2} \times 10}{0.1 \times 0.1}$$

$$V = \frac{10^{-1}}{10^{-2}} = 10\text{ V}$$

$$B_{\text{straight 1}} = \frac{\mu_0 i}{4\pi r} \otimes$$

$$B_{\text{semicircle}} = \frac{\mu_0 i}{4r} \otimes$$

$$B_{\text{straight 2}} = 0$$

$$B_{net} = B_{\text{straight 1}} + B_{\text{semicircle}} + B_{\text{straight 2}}$$

$$B_{net} = \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4r} + 0$$

$$B_{net} = \frac{\mu_0 i}{2r} \left( \frac{1}{2} + \frac{1}{2\pi} \right)$$

Assertion A: Electromagnets are made of soft iron. This is true — soft iron is the standard core material for electromagnets.

Reason R: Soft iron has high permeability and low retentivity.

High permeability means soft iron gets magnetized very easily when placed in a magnetic field. Low retentivity means it loses its magnetism quickly when the magnetizing field is removed.

These are exactly the properties needed for an electromagnet — it should magnetize strongly when current flows, and demagnetize immediately when current stops.

Both A and R are correct, and R correctly explains why soft iron is used for electromagnets (high permeability for strong magnetization, low retentivity for easy demagnetization).

Therefore, the answer is Option C.

The magnetisation of a linear magnetic medium is given by $$\mathbf{M} = \chi \mathbf{H}$$, where $$\chi$$ is the (volume) magnetic susceptibility and $$\mathbf{H}$$ is the magnetising field.

Relative permeability is defined as $$\mu_r = 1 + \chi$$ $$-(1)$$

Case 1 → Statement I
For a diamagnetic substance:
  • The induced dipoles are opposite to the applied field, hence $$\chi$$ is negative.
  • In naturally occurring materials $$|\chi|$$ is very small (of the order $$10^{-5}$$).
  • Condition $$\mu_r \gt 0$$ must hold for ordinary matter. From $$(1)$$ this gives
    $$1 + \chi \gt 0 \;\;\Longrightarrow\;\; \chi \gt -1$$

Combining the facts “$$\chi$$ is negative” and “$$\chi \gt -1$$” we obtain the range
$$-1 \le \chi \lt 0$$

Therefore Statement I is correct.

Case 2 → Statement II
The potential energy of a small sample in a non-uniform field is $$U = -\tfrac{1}{2}\,\chi V \mu_0 H^2$$, where $$V$$ is its volume. For a diamagnet $$\chi \lt 0$$, so $$U$$ increases with $$H^2$$. To minimise energy the sample drifts toward the region where $$H$$ is smaller, that is, from the stronger part of the field to the weaker part.

Hence Statement II is also correct.

Since both statements are true, the correct choice is Option D.

Statement I: If the number of turns N is doubled, current sensitivity = $$\frac{NBA}{k}$$. Doubling N doubles current sensitivity. This is true.

Statement II: Voltage sensitivity = Current sensitivity / Resistance = $$\frac{NBA}{kR}$$. When N doubles, resistance R also doubles (since $$R \propto N$$). So voltage sensitivity = $$\frac{2NBA}{k \cdot 2R} = \frac{NBA}{kR}$$, which remains the same. This is false.

The correct answer is Option 1: Statement I is true but Statement II is false.

For Statement I:

Diamagnetic susceptibility ($$\chi$$) is independent of temperature because the underlying induction mechanism does not rely on aligning pre-existing magnetic dipoles against thermal agitation. Therefore, Statement I is false.

For Statement II:

According to Lenz's law, when a substance is placed in an external magnetising field, the induced magnetic dipole moment always opposes the applied field. Therefore, Statement II is true.

$$I_s = \frac{NBA}{k}, \quad V_s = \frac{I_s}{R_g}$$

$$\text{Given that the resistance of the galvanometer coil } R_g \text{ is kept constant: } V_s \propto I_s$$

$$\% \Delta V_s = \% \Delta I_s = +25\%$$

Given: Cross-section area $$A = 2$$ cm$$^2$$ $$= 2 \times 10^{-4}$$ m$$^2$$, length $$L = 40$$ cm $$= 0.4$$ m, $$N = 400$$ turns, $$I = 0.4$$ A, total magnetic flux $$\Phi = 4\pi \times 10^{-6}$$ Wb.

The magnetic flux inside the solenoid is:

$$ \Phi = B \times A = \mu_r \mu_0 n I \times A $$

where $$n = N/L$$ is the number of turns per unit length.

$$ n = \frac{400}{0.4} = 1000 \text{ turns/m} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 1000 \times 0.4 \times 2 \times 10^{-4} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 400 \times 2 \times 10^{-4} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 8 \times 10^{-2} $$

$$ \Phi = \mu_r \times 32\pi \times 10^{-9} = \mu_r \times 3.2\pi \times 10^{-8} $$

Setting this equal to the given flux:

$$ \mu_r \times 3.2\pi \times 10^{-8} = 4\pi \times 10^{-6} $$

$$ \mu_r = \frac{4 \times 10^{-6}}{3.2 \times 10^{-8}} = \frac{4}{0.032} = 125 $$

Inside a solenoid, the magnetic field is uniform and directed along the axis.

When an electron moves with constant velocity along the axis of the solenoid, its velocity $$\vec{v}$$ is parallel to the magnetic field $$\vec{B}$$.

The magnetic force on the electron is:

$$\vec{F} = q(\vec{v} \times \vec{B})$$

Since $$\vec{v} \parallel \vec{B}$$, the cross product $$\vec{v} \times \vec{B} = 0$$.

Therefore:

(B) The electron will not experience magnetic force — True

(C) The electron will continue to move along the axis — True (no force to change its motion)

(A) Force along axis — False (no force at all)

(D) Accelerated along axis — False

(E) Parabolic path — False

The correct answer is B and C only.

We need to find the percentage increase in the magnetic field inside a toroid when the free space is filled with a material of magnetic susceptibility $$\chi = 2 \times 10^{-2}$$.

For a toroid with $$n$$ turns per unit length carrying current $$I$$, the magnetic field in free space is $$B_0 = \mu_0 n I$$.

When the interior is filled with a magnetic material, the relative permeability becomes $$\mu_r = 1 + \chi$$, so the total magnetic field is:

$$B = \mu_0(1 + \chi) n I = B_0(1 + \chi)$$

(This is because the magnetizing field $$H = nI$$ remains the same, being determined by the free current, and $$B = \mu H = \mu_0(1 + \chi)H$$.)

Now, the percentage increase in the magnetic field is:

$$\frac{\Delta B}{B_0} \times 100 = \frac{B_0 \chi}{B_0} \times 100 = \chi \times 100 = 2 \times 10^{-2} \times 100 = 2\%$$

Hence, the correct answer is Option 3.

$$T = 2\pi\sqrt{\frac{I}{M B_H}} \implies \nu = \frac{1}{2\pi}\sqrt{\frac{M B_H}{I}} \implies \nu \propto \sqrt{B_H} \implies \nu^2 \propto B \cos\theta$$

$$\frac{\nu_1^2}{\nu_2^2} = \frac{B_1 \cos\theta_1}{B_2 \cos\theta_2}$$

$$\left(\frac{20}{30}\right)^2 = \frac{B_1 \cos 30^\circ}{B_2 \cos 60^\circ} \implies \frac{4}{9} = \frac{B_1}{B_2} \times \frac{\sqrt{3}/2}{1/2} = \frac{B_1}{B_2}\sqrt{3}$$

$$\frac{B_1}{B_2} = \frac{4}{9\sqrt{3}}$$

$$\text{Comparing with the given ratio } \frac{4}{\sqrt{x}}: \sqrt{x} = 9\sqrt{3} = \sqrt{81 \times 3} = \sqrt{243} \implies x = 243$$

Given: Length $$l = 15$$ cm = 0.15 m, number of turns $$N = 60$$, magnetic intensity $$H = 2.4 \times 10^3$$ A/m.

The magnetic field intensity inside a solenoid is:

$$H = nI = \frac{N}{l}I$$

To demagnetise the bar magnet, the solenoid must produce a field of the same intensity:

$$I = \frac{Hl}{N} = \frac{2.4 \times 10^3 \times 0.15}{60} = \frac{360}{60} = 6 \text{ A}$$

The current required is 6 A.

We are given that the magnetic intensity at the centre of a long current-carrying solenoid is $$H = 1.6 \times 10^3$$ A/m, and the number of turns is $$n = 8$$ turns/cm.

Convert turns per cm to turns per metre: $$ n = 8 \text{ turns/cm} = 8 \times 100 = 800 \text{ turns/m} $$

Use the formula for magnetic intensity in a solenoid. For a long solenoid, the magnetic intensity inside is:

$$ H = nI $$

where $$n$$ is turns per unit length and $$I$$ is the current.

Solve for current: $$ I = \frac{H}{n} = \frac{1.6 \times 10^3}{800} = \frac{1600}{800} = 2 \text{ A} $$

The current flowing through the solenoid is 2 A.

A compass needle oscillates 20 times/min at place P (dip $$30^\circ$$) and 10 times/min at place Q (dip $$60^\circ$$). Find $$B_Q : B_P$$.

Recall that the frequency of oscillation of a magnetic needle is given by $$f = \frac{1}{2\pi}\sqrt{\frac{MB_H}{I}}$$, where $$B_H$$ is the horizontal component of the magnetic field, $$M$$ is the magnetic moment, and $$I$$ is the moment of inertia. Since $$f \propto \sqrt{B_H}$$, it follows that $$f^2 \propto B_H$$.

The horizontal component of the magnetic field relates to the total magnetic field by $$B_H = B\cos\delta$$, where $$\delta$$ is the angle of dip. Therefore, at place P, $$B_{H_P} = B_P \cos 30^\circ = B_P \cdot \frac{\sqrt{3}}{2}$$ and at place Q, $$B_{H_Q} = B_Q \cos 60^\circ = B_Q \cdot \frac{1}{2}$$.

Using the ratio of squared frequencies, we write $$\frac{f_P^2}{f_Q^2} = \frac{B_{H_P}}{B_{H_Q}}$$. Substituting $$f_P = 20$$ and $$f_Q = 10$$ yields $$\frac{20^2}{10^2} = \frac{B_P \cdot \frac{\sqrt{3}}{2}}{B_Q \cdot \frac{1}{2}}$$, which simplifies to $$4 = \frac{B_P \sqrt{3}}{B_Q}$$. From this it follows that $$\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$$.

Hence, the required ratio of the magnitudes of the magnetic fields is $$B_Q : B_P = \sqrt{3} : 4$$. Therefore, the answer is Option A: $$\sqrt{3} : 4$$.

We need to find the magnetic force $$F$$ acting on the segment $$CD$$ of the triangular wire loop placed in a uniform magnetic field.

The magnetic force on a straight current-carrying conductor of length $$L$$ carrying a current $$I$$ in a uniform magnetic field $$B$$ is given by the formula: $$F = I L B \sin\theta$$, where $$\theta$$ is the angle between the direction of the current element and the magnetic field vector.

From the given diagram, the triangular section $$\Delta BCD$$ is an equilateral triangle with sides $$BC = CD = BD = 5\text{ cm} = 0.05\text{ m}$$. Therefore, each internal angle of the triangle is $$60^\circ$$.

The magnetic field lines are directed horizontally from left to right, parallel to the segments $$AB$$ and $$DE$$. Therefore, the side $$BD$$ lies perfectly parallel to the magnetic field.

In the equilateral triangle, the side $$CD$$ slants downward towards the right. The angle between the direction of the current along segment $$CD$$ and the horizontal magnetic field lines is equal to the interior angle of the triangle at vertex $$D$$, which is $$\theta = 60^\circ$$.

We are given the following values:

Current, $$I = 10\text{ A}$$

Length of segment $$CD$$, $$L = 5\text{ cm} = 0.05\text{ m}$$

Magnetic field strength, $$B = 0.5\text{ T}$$

Angle, $$\theta = 60^\circ$$

Substituting these values into the magnetic force formula yields: $$F = 10 \times 0.05 \times 0.5 \times \sin(60^\circ)$$.

Since $$\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$$, the calculation becomes: $$F = 0.25 \times 0.866 = 0.2165\text{ N}$$.

Rounding to three decimal places, the magnitude of the force is approximately $$0.216\text{ N}$$.

Therefore, the correct answer is Option C: 0.216 N.

The magnetic susceptibility of the material inside the solenoid is $$\chi_m = 1.2 \times 10^{-5}$$.

Write the magnetic field with and without the magnetic material.

With air (vacuum): $$B_0 = \mu_0 nI$$

With the magnetic material: $$B = \mu_0(1 + \chi_m) nI$$

Calculate the fractional increase in magnetic field.

$$\frac{\Delta B}{B_0} = \frac{B - B_0}{B_0} = \frac{\mu_0(1 + \chi_m)nI - \mu_0 nI}{\mu_0 nI}$$

$$\frac{\Delta B}{B_0} = (1 + \chi_m) - 1 = \chi_m = 1.2 \times 10^{-5}$$

The fractional increase in the magnetic field is equal to the magnetic susceptibility, which is $$1.2 \times 10^{-5}$$.

The correct answer is Option A.

A cyclotron with magnetic field $$B = 1.0$$ T and radius $$R = 60$$ cm = 0.6 m accelerates protons. Find the kinetic energy in MeV.

Since for a charged particle in a cyclotron at the maximum radius the magnetic force equals the centripetal force, we have

$$qvB = \frac{mv^2}{R} \implies mv = qBR$$

From this relation one obtains

$$v = \frac{qBR}{m}$$

Substituting the expression for $$v$$ into the kinetic energy formula yields

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$$

Substituting $$q = e = 1.6 \times 10^{-19}$$ C, $$B = 1.0$$ T, $$R = 0.6$$ m, and $$m_p = 1.6 \times 10^{-27}$$ kg into this expression gives

$$KE = \frac{(1.6 \times 10^{-19})^2 \times (1.0)^2 \times (0.6)^2}{2 \times 1.6 \times 10^{-27}}$$

Numerator: $$(1.6)^2 \times 10^{-38} \times 0.36 = 2.56 \times 0.36 \times 10^{-38} = 0.9216 \times 10^{-38}$$

Denominator: $$3.2 \times 10^{-27}$$

Therefore,

$$KE = \frac{0.9216 \times 10^{-38}}{3.2 \times 10^{-27}} = 0.288 \times 10^{-11} = 2.88 \times 10^{-12} \text{ J}$$

Since $$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$ and $$1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$$,

$$KE = \frac{2.88 \times 10^{-12}}{1.6 \times 10^{-13}} = 18 \text{ MeV}$$

Answer: Option B: 18

We need to find the frequency of revolution of an electron moving at a right angle to the Earth's magnetic field.

1. Identify the Given Parameters

• Magnetic field ($$B$$) = $$1 \times 10^{-4}\text{ Wb m}^{-2}$$ (or Tesla, $$\text{T}$$)
• Mass of an electron ($$m$$) = $$9.0 \times 10^{-31}\text{ kg}$$
• Charge of an electron ($$e$$) = $$1.6 \times 10^{-19}\text{ C}$$
• Kinetic energy = $$0.1\text{ keV}$$ (Note: Since the electron moves at a right angle, its path is perfectly circular. The frequency of revolution in a magnetic field is independent of its speed or energy).

2. Formula for Frequency of Revolution

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the necessary centripetal force:

$$evB = \frac{mv^2}{r} \implies \frac{v}{r} = \frac{eB}{m}$$

Since the angular frequency is $$\omega = \frac{v}{r} = 2\pi f$$, the cyclotron frequency ($$f$$) is given by:

$$f = \frac{eB}{2\pi m}$$

3. Calculation

Substitute the given values into the frequency formula:

$$f = \frac{1.6 \times 10^{-19} \times 1 \times 10^{-4}}{2 \times 3.14 \times 9.0 \times 10^{-31}}$$

$$f = \frac{1.6 \times 10^{-23}}{56.52 \times 10^{-31}}$$

$$f \approx 0.0283 \times 10^8\text{ Hz} = 2.83 \times 10^6\text{ Hz}$$

Conclusion

The frequency of revolution of the electron is approximately $$2.8 \times 10^6\text{ Hz}$$, which matches Option C.

We need to determine the minimum current ($$I$$) required to keep a metallic rod stationary on a smooth inclined plane under the influence of a vertical magnetic field.

1. Identify the Given Parameters

• Linear mass density ($$\lambda = \frac{m}{l}$$) = $$0.45\text{ kg m}^{-1}$$
• Angle of inclination ($$\theta$$) = $$45^\circ$$
• Magnetic field strength ($$B$$) = $$0.15\text{ T}$$ (directed vertically upwards)
• Acceleration due to gravity ($$g$$) = $$10\text{ m s}^{-2}$$

2. Analyze the Forces Acting on the Rod

For the rod to remain stationary on the smooth inclined plane, the components of the forces acting parallel to the surface of the incline must perfectly balance each other:

• Gravitational Force Component: The component of the rod's weight acting down the incline is $$mg \sin\theta$$.
• Magnetic Force Component: A horizontal magnetic force ($$F_m = I l B$$) acts on the rod due to the vertical magnetic field. Resolving this force parallel to the incline, the component pushing up the incline is $$I l B \cos\theta$$.

3. Set Up the Equilibrium Equation

Equating the force component acting up the incline to the force component acting down the incline:

$$I l B \cos\theta = mg \sin\theta$$

Rearranging the formula to solve for the current ($$I$$):

$$I = \left(\frac{m}{l}\right) \frac{g}{B} \cdot \frac{\sin\theta}{\cos\theta}$$

$$I = \lambda \frac{g}{B} \tan\theta$$

4. Calculate the Minimum Current

Substitute the given numerical values into our derived equilibrium equation:

$$I = 0.45 \times \frac{10}{0.15} \times \tan(45^\circ)$$

Since $$\tan(45^\circ) = 1$$:

$$I = \frac{4.5}{0.15} \times 1 = 30\text{ A}$$

Therefore, the minimum current flowing in the rod required to keep it stationary is 30 A, which corresponds to Option A.

Two bar magnets oscillate in Earth's magnetic field with time periods $$T_1 = 3$$ s and $$T_2 = 4$$ s, and their moments of inertia are in the ratio $$I_1 : I_2 = 3 : 2$$. We need the ratio of their magnetic moments.

Since the time period of a bar magnet oscillating in a magnetic field $$B$$ is given by $$T = 2\pi\sqrt{\frac{I}{MB}},$$ where $$I$$ is the moment of inertia and $$M$$ is the magnetic moment, squaring this expression yields $$T^2 = 4\pi^2 \frac{I}{MB} \implies M = \frac{4\pi^2 I}{T^2 B}.$$

Therefore, the ratio of the magnetic moments becomes $$\frac{M_1}{M_2} = \frac{I_1}{I_2} \times \frac{T_2^2}{T_1^2}.$$

Substituting the given ratios into the above expression yields $$\frac{M_1}{M_2} = \frac{3}{2} \times \frac{16}{9} = \frac{48}{18} = \frac{8}{3}.$$

Hence, the correct answer is Option B: $$8:3$$.

Two concentric circular loops of radii $$r_1 = 30 \text{ cm}$$ and $$r_2 = 50 \text{ cm}$$ in the X-Y plane carry current $$I = 7 \text{ A}$$. We need to find the net magnetic moment.

The magnetic moment of a circular loop is:

$$\vec{M} = I \cdot A \cdot \hat{n}$$

where $$A = \pi r^2$$ is the area and $$\hat{n}$$ is the unit normal determined by the direction of current (right-hand rule).

From the figure, the inner loop ($$r_1 = 0.3 \text{ m}$$) carries current in one direction and the outer loop ($$r_2 = 0.5 \text{ m}$$) carries current in the opposite direction.

For the inner loop (current gives $$+\hat{k}$$):

$$M_1 = 7 \times \pi \times (0.3)^2 = 7 \times 0.09\pi = 0.63\pi \text{ A m}^2$$

For the outer loop (current gives $$-\hat{k}$$):

$$M_2 = 7 \times \pi \times (0.5)^2 = 7 \times 0.25\pi = 1.75\pi \text{ A m}^2$$

$$\vec{M}_{net} = 0.63\pi \hat{k} - 1.75\pi \hat{k} = -1.12\pi \hat{k}$$

$$\vec{M}_{net} = -1.12 \times 3.14 \hat{k} \approx -3.52 \hat{k} \approx -\dfrac{7}{2} \hat{k} \text{ A m}^2$$

The correct answer is Option B: $$-\dfrac{7}{2} \hat{k} \text{ A m}^2$$.

A velocity selector has $$\vec{E} = E\hat{k}$$ and $$\vec{B} = B\hat{j}$$ with $$B = 12 \text{ mT}$$. An electron has energy $$728 \text{ eV}$$ and moves along the positive x-axis. We need to find $$E$$.

The kinetic energy of the electron is $$KE = \dfrac{1}{2}mv^2 = 728 \text{ eV} = 728 \times 1.6 \times 10^{-19} \text{ J}$$, which simplifies to $$1164.8 \times 10^{-19} \text{ J} = 1.1648 \times 10^{-16} \text{ J}$$.

Substituting this value into the expression for velocity gives $$v^2 = \dfrac{2 \times KE}{m} = \dfrac{2 \times 1.1648 \times 10^{-16}}{9.1 \times 10^{-31}} = 2.56 \times 10^{14}$$, so that $$v = 1.6 \times 10^7 \text{ m s}^{-1}$$.

For the electron to pass undeflected through the selector, the electric force must balance the magnetic force, thus $$qE = qvB$$ and hence $$E = vB$$.

Substituting $$v = 1.6 \times 10^7 \text{ m s}^{-1}$$ and $$B = 12 \times 10^{-3}$$ yields $$E = 1.6 \times 10^7 \times 12 \times 10^{-3} = 19.2 \times 10^4 = 1.92 \times 10^5 \text{ V m}^{-1}$$, which is equal to $$192 \text{ kV m}^{-1}$$.

Therefore, the correct answer is Option A: $$192 \text{ kV m}^{-1}$$.

The angle of dip is $$\delta = 30°$$ and the horizontal component of Earth’s magnetic field is $$B_H = 0.5$$ G. These quantities are linked by the relation $$B_H = B\cos\delta$$, where $$B$$ denotes the total magnetic field. Solving for $$B$$ gives

Therefore, the correct answer is Option A: $$\frac{1}{\sqrt{3}}$$.

Given: Susceptibility $$\chi = 99$$, and $$\mu_0 = 4\pi \times 10^{-7}$$ Wb/A-m.

The relationship between permeability and susceptibility is:

$$\mu = \mu_0(1 + \chi)$$

Substituting the values:

$$\mu = 4\pi \times 10^{-7} \times (1 + 99)$$

$$\mu = 4\pi \times 10^{-7} \times 100$$

$$\mu = 4\pi \times 10^{-5} \text{ Wb/A-m}$$

The correct answer is Option A.

We have the vertical component of the Earth's magnetic field $$B_V = 6 \times 10^{-5}$$ T and the angle of dip $$\delta = 37°$$, with $$\tan 37° = \frac{3}{4}$$.

The relationship between the vertical component, horizontal component, and the resultant magnetic field is: $$B_V = B \sin \delta$$, where $$B$$ is the resultant (total) magnetic field.

We also know that $$\sin 37° = \frac{3}{5}$$ (since $$\tan 37° = 3/4$$, we have a 3-4-5 triangle giving $$\sin 37° = 3/5$$).

Therefore: $$B = \frac{B_V}{\sin 37°} = \frac{6 \times 10^{-5}}{3/5} = 6 \times 10^{-5} \times \frac{5}{3} = 10 \times 10^{-5} = 1 \times 10^{-4}$$ T.

Hence, the correct answer is Option 4.

A charged particle in a cyclotron has its kinetic energy increased to 4 times its original value, and we seek the ratio of the new radius to the original radius. In a magnetic field, the radius of the circular path of a charged particle is given by:

$$r = \frac{mv}{qB}$$

Since $$KE = \frac{1}{2}mv^2$$, it follows that $$v = \sqrt{\frac{2KE}{m}}$$. Substituting this result into the expression for the radius, one obtains:

$$r = \frac{m}{qB}\sqrt{\frac{2KE}{m}} = \frac{\sqrt{2mKE}}{qB}$$, which implies that $$r \propto \sqrt{KE}$$.

With the kinetic energy increased by a factor of 4, the ratio of the new radius to the original radius becomes: $$\frac{r_{\text{new}}}{r_{\text{original}}} = \sqrt{\frac{KE_{\text{new}}}{KE_{\text{original}}}} = \sqrt{\frac{4KE}{KE}} = \sqrt{4} = 2$$, so that the new radius is twice the original radius, or $$2 : 1$$.

Thus, the correct answer is Option C: 2 : 1.

An electromagnet needs to be easily magnetized when current flows and easily demagnetized when the current is switched off.

Understand the required properties: High permeability: Permeability measures how easily a material can be magnetized. A material with high permeability allows magnetic field lines to pass through it easily and becomes strongly magnetized even with a small applied magnetic field. This is essential for an electromagnet to be effective.

Low retentivity: Retentivity (or remanence) is the ability of a material to retain magnetization after the external magnetic field is removed. For an electromagnet, we need LOW retentivity so that the magnet can be easily switched off (demagnetized) when the current stops.

Low coercivity: Coercivity is the reverse field needed to demagnetize the material. Low coercivity also helps in easy demagnetization.

Evaluate the options: Option A: Low coercivity and high retentivity - High retentivity is undesirable as the electromagnet would remain magnetized after switching off.

Option B: Low coercivity and low permeability - Low permeability means the material would not magnetize strongly, making a weak electromagnet.

Option C: High permeability and low retentivity - This is the ideal combination: strong magnetization when ON, easy demagnetization when OFF.

Option D: High permeability and high retentivity - High retentivity would keep the magnet magnetized even after switching off.

Soft iron has high permeability (easily magnetized) and low retentivity (easily demagnetized), making it ideal for electromagnets.

The correct answer is Option C.

Two long parallel conductors are 8 cm apart. The magnetic field at the midpoint is 300 $$\mu$$T. Find the equal current and direction.

The midpoint is at 4 cm = 0.04 m from each conductor.

Magnetic field due to each conductor at the midpoint:

$$B = \frac{\mu_0 I}{2\pi d} = \frac{4\pi \times 10^{-7} \times I}{2\pi \times 0.04} = \frac{2 \times 10^{-7} \times I}{0.04} = 5 \times 10^{-6} I \text{ T}$$

If currents are in the same direction: At the midpoint, the fields due to the two wires point in opposite directions (by right-hand rule). The net field = 0. This contradicts 300 $$\mu$$T.

If currents are in opposite directions: At the midpoint, the fields add up:

$$B_{\text{net}} = 2B = 2 \times 5 \times 10^{-6} I = 10^{-5} I$$

$$300 \times 10^{-6} = 10^{-5} I$$

$$I = \frac{300 \times 10^{-6}}{10^{-5}} = 30 \text{ A}$$

Hence, the correct answer is Option C: 30 A in the opposite direction.

The magnetic field at the center of a closely wound circular coil is given by the formula:

$$ B = \frac{\mu_0 N I}{2R} $$

Here, $$ B $$ is the magnetic field, $$ \mu_0 $$ is the permeability of free space ($$ 4\pi \times 10^{-7} $$ T m/A), $$ N $$ is the number of turns, $$ I $$ is the current, and $$ R $$ is the radius of the coil.

We are given:

• Radius $$ R = 5 $$ cm = 0.05 m (converted to meters for SI units),
• Magnetic field $$ B = 37.68 \times 10^{-4} $$ T,
• Number of turns $$ N = 100 $$,
• $$ \pi = 3.14 $$.

First, express $$ \mu_0 $$ using the given $$ \pi $$:

$$ \mu_0 = 4 \pi \times 10^{-7} = 4 \times 3.14 \times 10^{-7} = 12.56 \times 10^{-7} = 1.256 \times 10^{-6} \text{ T m/A} $$

Now, rearrange the formula to solve for $$ I $$:

$$ I = \frac{2 R B}{\mu_0 N} $$

Substitute the known values:

$$ I = \frac{2 \times 0.05 \times 37.68 \times 10^{-4}}{1.256 \times 10^{-6} \times 100} $$

Compute the numerator:

$$ 2 \times 0.05 = 0.1 $$

$$ 0.1 \times 37.68 \times 10^{-4} = 0.1 \times 0.003768 = 0.0003768 = 3.768 \times 10^{-4} $$

Compute the denominator:

$$ 1.256 \times 10^{-6} \times 100 = 1.256 \times 10^{-4} $$

Now, divide the numerator by the denominator:

$$ I = \frac{3.768 \times 10^{-4}}{1.256 \times 10^{-4}} = \frac{3.768}{1.256} $$

Perform the division:

$$ 3.768 \div 1.256 = 3 $$

Therefore, the current $$ I $$ is 3 A.

Hence, the correct answer is 3.

The magnetic field at any point is obtained by adding the contributions from each current-carrying segment separately.
For every straight segment lying in the $$xy$$-plane the field at the point O is perpendicular to the plane, i.e. along $$\pm \hat{k}$$. We therefore have to calculate only the magnitudes and check the sign with the right-hand rule.

Magnetic field of a straight wire.
For an infinitely long straight wire at a perpendicular distance $$d$$ from the observation point, $$B=\dfrac{\mu_0 I}{2\pi d}$$.

Magnetic field of a finite straight wire.
For a straight piece of length $$AB$$ the field at a point P lying at a perpendicular distance $$r$$ from the wire is $$B=\dfrac{\mu_0 I}{4\pi r}\left(\sin\theta_1+\sin\theta_2\right)$$ $$-(1)$$ where $$\theta_1$$ and $$\theta_2$$ are the angles made by the lines PA and PB with the perpendicular from P to the wire.

The geometry given in the figure consists of two parts: (i) an infinitely long wire whose closest distance from O is $$d=\dfrac{L}{2\pi}$$, and (ii) a straight segment of length $$L$$ lying parallel to the $$x$$-axis at a height $$y=L$$ above O, its right end being at $$(L,L)$$. Both carry the same current $$I$$.

(i) Field due to the infinitely long wire
Using the expression written above, $$B_1=\dfrac{\mu_0 I}{2\pi d}=\dfrac{\mu_0 I}{2\pi\left(L/2\pi\right)} =\dfrac{\mu_0 I}{L}.$$ With the current direction shown in the figure, the right-hand rule gives the direction as $$-\hat{k}$$.

(ii) Field due to the finite segment of length $$L$$
For this segment the perpendicular distance from O is $$r=L$$. The nearer end of the segment lies directly above O, so $$\theta_1=0^{\circ}$$ and hence $$\sin\theta_1=0$$. For the farther end at $$(L,L)$$, the line joining it to O makes an angle of $$45^{\circ}$$ with the perpendicular; therefore $$\sin\theta_2=\sin45^{\circ}=\dfrac{1}{\sqrt2}$$. Substituting these values in $$(1)$$, $$B_2=\dfrac{\mu_0 I}{4\pi L}\left(0+\dfrac{1}{\sqrt2}\right) =\dfrac{\mu_0 I}{4\sqrt2\pi L}.$$ The same right-hand rule shows that this field is also along $$-\hat{k}$$.

Net magnetic field at O
Both contributions point along $$-\hat{k}$$, so the magnitudes simply add: $$\vec{B}= -\hat{k}\left(\dfrac{\mu_0 I}{L} +\dfrac{\mu_0 I}{4\sqrt2\pi L}\right) =\dfrac{-\mu_0 I}{L}\left(1+\dfrac{1}{4\sqrt2\pi}\right)\hat{k}.$$

Comparing with the given options, this is exactly Option C.

Option C which is: $$\vec{B} = \dfrac{-\mu_0 I}{L}\left(1 + \dfrac{1}{4\sqrt{2}\pi}\right)\hat{k}.$$

In a ferromagnetic material below the Curie temperature, the material is divided into small regions called magnetic domains. Within each domain, the atomic magnetic dipoles are aligned parallel to each other due to strong exchange interactions between neighbouring atoms.

This parallel alignment of all magnetic dipoles within a domain means that each domain is magnetised to saturation, i.e., it has the maximum possible magnetisation. The domain as a whole behaves like a tiny magnet with a definite magnetisation direction.

However, different domains may have their magnetisation directions oriented differently, so the net magnetisation of the bulk material can be zero in the demagnetised state. When an external magnetic field is applied, domains aligned with the field grow at the expense of others, leading to net magnetisation of the material.

Therefore, a domain in a ferromagnetic material below the Curie temperature is defined as a macroscopic region with saturation magnetisation.

The magnetic force on a moving charge is given by $$\vec{F} = Q(\vec{v} \times \vec{B})$$. By the properties of the cross product, this force is always perpendicular to the velocity $$\vec{v}$$ of the charge.

The work done by a force is $$dW = \vec{F} \cdot d\vec{l}$$, where $$d\vec{l}$$ is along the direction of displacement (and hence along $$\vec{v}$$). Since $$\vec{F} \perp \vec{v}$$, the dot product $$\vec{F} \cdot d\vec{l} = 0$$ at every instant.

Therefore, the work done by the magnetic field $$\vec{B}$$ on the charge is zero.

All three particles — proton, deuteron, and $$\alpha$$ particle — move with the same momentum $$p$$ in a uniform magnetic field $$B$$.

The magnetic force on a charged particle is $$F = qvB$$. Since $$p = mv$$, we have $$v = \frac{p}{m}$$, so $$F = \frac{qpB}{m}$$. The force ratio depends on $$\frac{q}{m}$$ for each particle.

For a proton: $$q_p = e$$, $$m_p = m$$, so $$\frac{q_p}{m_p} = \frac{e}{m}$$.

For a deuteron: $$q_d = e$$, $$m_d = 2m$$, so $$\frac{q_d}{m_d} = \frac{e}{2m}$$.

For an $$\alpha$$ particle: $$q_\alpha = 2e$$, $$m_\alpha = 4m$$, so $$\frac{q_\alpha}{m_\alpha} = \frac{2e}{4m} = \frac{e}{2m}$$.

The ratio of magnetic forces is $$F_p : F_d : F_\alpha = \frac{e}{m} : \frac{e}{2m} : \frac{e}{2m} = 2 : 1 : 1$$.

The speed of each particle is $$v = \frac{p}{m}$$, so the speed ratio is $$v_p : v_d : v_\alpha = \frac{1}{m} : \frac{1}{2m} : \frac{1}{4m} = 4 : 2 : 1$$.

Therefore the ratio of forces is $$2 : 1 : 1$$ and the speeds are in the ratio $$4 : 2 : 1$$, which corresponds to Option (1).

In a ferromagnetic material, the atomic magnetic dipoles are grouped into regions called magnetic domains. Within each domain, the dipoles are aligned in the same direction, but different domains may point in different directions.

When a soft ferromagnetic material is placed in an external magnetic field, the domains respond in two ways. First, the domains whose magnetic moments are aligned (or nearly aligned) with the external field tend to grow in size at the expense of neighbouring domains that are less favourably oriented. This is called domain wall motion. Second, the magnetic moments of individual domains may rotate to align more closely with the applied field direction. This is called domain rotation.

Depending on the strength and direction of the external field relative to each domain's initial orientation, some domains increase in size while others decrease. Additionally, the orientation of domains changes to align with the field. In a soft ferromagnetic material, both these processes occur readily because the domain walls move easily.

Therefore, the magnetic domains may increase or decrease in size and change their orientation.

The correct answer is that domains may increase or decrease in size and change its orientation.

The magnetic field at a point on the axis of a circular coil of radius $$R$$ at a distance $$x$$ from the centre is given by $$B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$$.

Given that the fields at distances $$x_1 = 0.05$$ m and $$x_2 = 0.2$$ m from the centre are in the ratio $$8 : 1$$, we have:

$$\frac{B_1}{B_2} = \frac{(R^2 + x_2^2)^{3/2}}{(R^2 + x_1^2)^{3/2}} = 8$$

Taking the cube root of both sides (since $$8^{1/3} = 2$$) and then squaring:

$$\frac{R^2 + x_2^2}{R^2 + x_1^2} = 8^{2/3} = 4$$

Substituting $$x_1 = 0.05$$ m and $$x_2 = 0.2$$ m:

$$R^2 + (0.2)^2 = 4(R^2 + (0.05)^2)$$

$$R^2 + 0.04 = 4R^2 + 0.01$$

$$0.03 = 3R^2$$

$$R^2 = 0.01$$

$$R = 0.1$$ m

The correct answer is Option (3): 0.1 m.

We need the magnetic field produced by a circular coil carrying a steady current. The standard expression for the magnitude of the magnetic field on the axis of a single loop of radius $$a$$, at a point whose distance from the centre measured along the axis is $$r$$, is

$$B(r)=\frac{\mu_{0}\,I\,a^{2}}{2\left(a^{2}+r^{2}\right)^{3/2}}.$$

(This formula can be derived from the Biot-Savart law, but we take it here as a known result.) At the centre of the same coil we simply place $$r=0$$, giving

$$B_{\text{centre}}=B(0)=\frac{\mu_{0}\,I\,a^{2}}{2\left(a^{2}+0\right)^{3/2}} =\frac{\mu_{0}\,I\,a^{2}}{2a^{3}} =\frac{\mu_{0}\,I}{2a}.$$

We now form the ratio of the field at the off-axis point to the field at the centre:

$$\frac{B(r)}{B_{\text{centre}}} =\frac{\dfrac{\mu_{0}I a^{2}}{2\left(a^{2}+r^{2}\right)^{3/2}}} {\dfrac{\mu_{0}I}{2a}} =\frac{a^{3}}{\left(a^{2}+r^{2}\right)^{3/2}} =\frac{1}{\left(1+\dfrac{r^{2}}{a^{2}}\right)^{3/2}}.$$

To find the fractional change we write

$$\text{Fractional change}=\frac{B(r)-B_{\text{centre}}}{B_{\text{centre}}} =\frac{B(r)}{B_{\text{centre}}}-1 =\left(1+\frac{r^{2}}{a^{2}}\right)^{-3/2}-1.$$

The question specifies $$r<a$$, so the quantity $$\dfrac{r^{2}}{a^{2}}$$ is small. We therefore expand the power using the binomial approximation. For a small $$x$$, the expansion

$$(1+x)^{n}\;\approx\;1+nx$$

is adequate to first order. Here $$n=-\dfrac{3}{2}$$ and $$x=\dfrac{r^{2}}{a^{2}}$$, so

$$\left(1+\frac{r^{2}}{a^{2}}\right)^{-3/2} \approx 1-\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Substituting this approximation into the expression for the fractional change gives

$$\text{Fractional change} \approx\Bigl[1-\frac{3}{2}\frac{r^{2}}{a^{2}}\Bigr]-1 =-\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

The negative sign merely tells us that the magnetic field at the point on the axis is smaller than the field at the centre. The magnitude of the fractional change is therefore

$$\left|\frac{B(r)-B_{\text{centre}}}{B_{\text{centre}}}\right| =\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Among the given options, this matches

$$\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Hence, the correct answer is Option C.

A bar magnet of length $$2l = 14$$ cm (so $$l = 7$$ cm $$= 0.07$$ m) is placed in the magnetic meridian with its north pole pointing towards geographic north. A neutral point is found at a distance $$r = 18$$ cm $$= 0.18$$ m from the center of the magnet.

When the north pole of the magnet points toward geographic north, the magnet's axial field reinforces Earth's horizontal field on both sides along the axis. On the equatorial line (perpendicular bisector), the magnet's field points opposite to the magnet's magnetic moment direction (i.e., southward), opposing the Earth's field. So the neutral points lie on the equatorial line.

At the neutral point on the equatorial line, the magnetic field due to the magnet equals $$B_H$$: $$\frac{\mu_0}{4\pi} \cdot \frac{M}{(r^2 + l^2)^{3/2}} = B_H$$.

Solving for $$M$$: $$M = \frac{B_H \cdot (r^2 + l^2)^{3/2}}{\mu_0 / (4\pi)}$$.

Computing $$r^2 + l^2 = (0.18)^2 + (0.07)^2 = 0.0324 + 0.0049 = 0.0373$$ m$$^2$$.

$$(r^2 + l^2)^{3/2} = (0.0373)^{3/2} = 0.0373 \times \sqrt{0.0373} = 0.0373 \times 0.19314 = 7.204 \times 10^{-3}$$ m$$^3$$.

With $$B_H = 0.4$$ G $$= 4 \times 10^{-5}$$ T and $$\frac{\mu_0}{4\pi} = 10^{-7}$$ T m/A:

$$M = \frac{4 \times 10^{-5} \times 7.204 \times 10^{-3}}{10^{-7}} = \frac{2.8816 \times 10^{-7}}{10^{-7}} = 2.880$$ J T$$^{-1}$$.

For a charged particle moving in a magnetic field, the radius of the circular path is $$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$$, where $$K$$ is the kinetic energy.

For the deuteron: mass $$m_d = 2u$$, charge $$q_d = e$$. For the alpha particle: mass $$m_\alpha = 4u$$, charge $$q_\alpha = 2e$$.

Both particles have equal kinetic energy $$K$$ and enter the same magnetic field $$B$$. Taking the ratio:

$$\frac{r_d}{r_\alpha} = \frac{\sqrt{2m_d K}/(q_d B)}{\sqrt{2m_\alpha K}/(q_\alpha B)} = \frac{q_\alpha}{q_d} \cdot \sqrt{\frac{m_d}{m_\alpha}} = \frac{2e}{e} \cdot \sqrt{\frac{2u}{4u}} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$$

The radius of the circular trajectory of a charged particle in a magnetic field is $$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$$, where $$K$$ is the kinetic energy.

For a proton: $$r_p = \frac{\sqrt{2m_p K_p}}{eB}$$. For an $$\alpha$$-particle (mass $$4m_p$$, charge $$2e$$): $$r_\alpha = \frac{\sqrt{2 \cdot 4m_p \cdot K_\alpha}}{2eB} = \frac{\sqrt{8m_p K_\alpha}}{2eB}$$.

The ratio is: $$\frac{r_p}{r_\alpha} = \frac{\sqrt{2m_p K_p}}{eB} \cdot \frac{2eB}{\sqrt{8m_p K_\alpha}} = \frac{2\sqrt{K_p}}{2\sqrt{K_\alpha}} = \sqrt{\frac{K_p}{K_\alpha}}$$.

Given $$\frac{r_p}{r_\alpha} = 2$$, we get $$\frac{K_p}{K_\alpha} = 4$$, so $$K_p : K_\alpha = 4 : 1$$.

The magnetic flux density inside a solenoid with a core of relative permeability $$\mu_r$$ is given by $$B = \mu_r \mu_0 n I$$, where $$n$$ is the number of turns per metre and $$I$$ is the current.

Substituting the given values: $$B = 500 \times 4\pi \times 10^{-7} \times 1000 \times 5$$. Computing step by step: $$4\pi \times 10^{-7} \times 1000 = 4\pi \times 10^{-4}$$, then $$500 \times 4\pi \times 10^{-4} = 2000\pi \times 10^{-4} = 0.2\pi$$, and finally $$0.2\pi \times 5 = \pi$$ T.

The correct answer is option 1: $$\pi$$ T.

We have two ions that both enter a region where the magnetic field $$\vec B$$ is uniform and perpendicular to their velocities. Whenever a charged particle moves perpendicular to a magnetic field, it describes a circular path. The required formula for the radius of that circular path is first recalled:

For a particle of mass $$m$$, charge $$q$$ and speed $$v$$ in a magnetic field of magnitude $$B$$ (with velocity perpendicular to the field), the magnetic force provides the necessary centripetal force. Hence

$$q\,v\,B = \dfrac{m v^{2}}{r}.$$

Solving this equation for the radius $$r$$ gives

$$r = \dfrac{m\,v}{q\,B} \;.$$

In the present question, both ions possess the same kinetic energy. Let us denote that common kinetic energy by $$K$$. The kinetic energy of any particle is related to its speed by the well-known relation

$$K = \dfrac{1}{2} m v^{2}.$$

From this we can express the speed $$v$$ of each ion in terms of its mass and the common kinetic energy:

$$v = \sqrt{\dfrac{2K}{m}}.$$

Now we substitute this expression for $$v$$ into the radius formula. Performing the substitution, we get

$$r = \dfrac{m}{q\,B}\;\sqrt{\dfrac{2K}{m}}.$$

By taking the square root of $$m$$ in the numerator, the algebraic simplification proceeds as follows:

$$r = \dfrac{\sqrt{m}\;\sqrt{m}}{q\,B}\;\sqrt{\dfrac{2K}{m}} = \dfrac{\sqrt{m}\;}{q\,B}\;\sqrt{2K} = \dfrac{\sqrt{2K}}{B}\;\dfrac{\sqrt{m}}{q}.$$

The magnetic field $$B$$ and the factor $$\sqrt{2K}$$ are identical for both ions, so the only quantity that decides the relative radii is the ratio $$\dfrac{\sqrt{m}}{q}$$.

Let us calculate this ratio for each ion separately.

Ion 1 (lighter ion)
Mass $$m_{1}=4\ \text{amu},\qquad q_{1}=+2e.$$

$$\dfrac{\sqrt{m_{1}}}{q_{1}} = \dfrac{\sqrt{4}}{2e} = \dfrac{2}{2e} = \dfrac{1}{e}.$$

Ion 2 (heavier ion)
Mass $$m_{2}=16\ \text{amu},\qquad q_{2}=+3e.$$

$$\dfrac{\sqrt{m_{2}}}{q_{2}} = \dfrac{\sqrt{16}}{3e} = \dfrac{4}{3e}.$$

We now compare these two numerical factors:

$$\dfrac{\sqrt{m_{2}}}{q_{2}} = \dfrac{4}{3e}\;>\;\dfrac{1}{e} = \dfrac{\sqrt{m_{1}}}{q_{1}}.$$

The radius of curvature is directly proportional to this factor, so

$$r_{2} > r_{1}.$$

Because the heavier ion (Ion 2) travels in a circle of larger radius, it is bent less sharply by the magnetic field. Conversely, the lighter ion (Ion 1) follows a circle of smaller radius, meaning its trajectory bends more.

In ordinary language, a smaller radius corresponds to greater deflection; a larger radius corresponds to less deflection. Therefore the lighter ion is deflected more, and the heavier ion is deflected less.

Hence, the correct answer is Option B.

We have a long straight coaxial system. Its inner solid conductor has radius $$a$$ and carries a total current $$I_0$$ uniformly over its cross-section. Its outer cylindrical shell occupies the space from $$r=b$$ to $$r=c$$ and carries an equal current $$I_0$$ but in the opposite direction, also distributed uniformly through its own metal. Because the currents are steady and the geometry is perfectly coaxial, we take every magnetic field line to be a circle centred on the common axis, so the magnitude of the field depends only on the radial distance $$r$$ from that axis.

For such circular symmetry we apply Ampère’s circuital law, first stating it explicitly:

$$\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enclosed}}.$$

Choosing a circular Amperian path of radius $$r$$ (so $$d\vec l$$ is everywhere tangential and of magnitude $$r\,d\theta$$), the left side becomes $$B(2\pi r)$$. Hence

$$B(2\pi r)=\mu_0 I_{\text{enclosed}}\quad\Longrightarrow\quad B=\dfrac{\mu_0 I_{\text{enclosed}}}{2\pi r}.$$

We now find the enclosed current in the two separate radial zones asked for and then form their ratio.

Case (i)  $$r=x\lt a$$

The current density $$J$$ in the inner wire is uniform, so

$$J=\dfrac{I_0}{\text{cross-sectional area}}=\dfrac{I_0}{\pi a^2}.$$

The area enclosed by our path of radius $$x$$ is $$\pi x^2$$, hence the enclosed current is

$$I_{\text{enc}}^{(i)} = J(\pi x^2)=\dfrac{I_0}{\pi a^2}\,(\pi x^2)=I_0\,\dfrac{x^2}{a^2}.$$

Substituting this into Ampère’s result gives

$$B_{(i)}=\dfrac{\mu_0}{2\pi x}\,I_0\,\dfrac{x^2}{a^2}=\dfrac{\mu_0 I_0\,x}{2\pi a^2}.$$

Case (ii)  $$a\lt r=x\lt b$$

Now the entire inner conductor lies inside the Amperian loop, while none of the outer shell (which starts only at $$r=b$$) is enclosed. Therefore

$$I_{\text{enc}}^{(ii)}=I_0.$$

Inserting this into Ampère’s expression, we find

$$B_{(ii)}=\dfrac{\mu_0 I_0}{2\pi x}.$$

Forming the required ratio

$$\dfrac{B_{(i)}}{B_{(ii)}}=\dfrac{\dfrac{\mu_0 I_0\,x}{2\pi a^2}}{\dfrac{\mu_0 I_0}{2\pi x}} =\dfrac{x^2}{a^2}.$$

This matches Option D in the given list.

Hence, the correct answer is Option D.

First recall the magnetic-field formula for a straight, finite conductor. For a wire segment of length $$L$$ that carries current $$I$$, the field at a point that is at a perpendicular distance $$r$$ from the wire and that “sees’’ the two ends under angles $$\theta_1$$ and $$\theta_2$$ (measured from the perpendicular) is

$$ B=\dfrac{\mu_0 I}{4\pi r}\,(\sin\theta_1+\sin\theta_2), $$

and the direction is obtained from the right-hand rule for the element $$I\,\vec{\mathrm d\ell}\times\vec r$$.

The triangle is equilateral with side

$$ a = 9\ \text{cm}=0.09\ \text{m}, $$

and the current is $$I = 1.5\ \text{A}$$, flowing clockwise. The point at which we need the field is the centroid $$G$$. In an equilateral triangle the perpendicular distance from the centroid to any side is one third of the altitude. The altitude is

$$ h = a\dfrac{\sqrt{3}}{2}, $$

so the required distance is

$$ r = \dfrac{h}{3}= \dfrac{a}{3}\,\dfrac{\sqrt{3}}{2} = a\,\dfrac{\sqrt{3}}{6} = 0.09\ \text{m}\times\dfrac{\sqrt{3}}{6}. $$

Numerically, using $$\sqrt{3}=1.732$$,

$$ r = 0.09\times\dfrac{1.732}{6} = 0.09\times0.2887 \approx 0.02598\ \text{m}. $$

Because the centroid lies on the perpendicular bisector of every side, the two angles for each side are equal: $$\theta_1=\theta_2=\theta$$. The half-length of a side is $$a/2$$, so

$$ \tan\theta = \dfrac{a/2}{r} = \dfrac{0.09/2}{0.02598} = \dfrac{0.045}{0.02598} \approx 1.732 =\sqrt{3}. $$

Hence $$\theta=60^\circ$$ and

$$ \sin\theta = \sin60^\circ = \dfrac{\sqrt{3}}{2}. $$

Substituting $$\theta_1=\theta_2=60^\circ$$ in the straight-wire formula gives the field from one side:

$$ B_1 = \dfrac{\mu_0 I}{4\pi r}\, (\sin\theta+\sin\theta) = \dfrac{\mu_0 I}{4\pi r}\,(2\sin60^\circ) = \dfrac{\mu_0 I}{4\pi r}\, \left(2\times\dfrac{\sqrt{3}}{2}\right) = \dfrac{\mu_0 I\sqrt{3}}{4\pi r}. $$

Now insert the numerical values. We have $$\mu_0/4\pi = 10^{-7}\,{\rm T\,m\,A^{-1}}$$, so

$$ B_1 = 10^{-7} \times\dfrac{1.5\sqrt{3}}{0.02598} = 10^{-7} \times\dfrac{1.5\times1.732}{0.02598}. $$

The numerator is $$1.5\times1.732 = 2.598$$. Dividing,

$$ \dfrac{2.598}{0.02598}\approx 100.0. $$

Therefore

$$ B_1 \approx 100.0\times10^{-7}\ \text{T} = 1.0\times10^{-5}\ \text{T}. $$

The triangle has three identical sides, and by symmetry each contributes the same magnitude and the same axial direction, so the net field is

$$ B = 3B_1 = 3\times1.0\times10^{-5}\ \text{T} = 3.0\times10^{-5}\ \text{T}. $$

Finally, we determine the sense (into or out of the plane). Curl the fingers of your right hand in the direction of the current (clockwise as viewed from the front); the thumb then points into the page, that is, inside the plane of the triangle. All three side contributions add with this same orientation, so the resultant field is into the plane.

Hence, the correct answer is Option D.

Upon reviewing the physical definitions carefully, the previous derivation holds perfectly true, and the final solution is completely accurate. Let's do a quick structural review of the properties to reinforce why this is the correct answer:

Properties Review for Diamagnetic Materials

• Magnetization ($$M$$) vs. Magnetising Field ($$H$$): Since the magnetic susceptibility ($$\chi$$) of a diamagnetic material is a small negative constant, the relationship $$M = \chi H$$ yields a straight line with a negative slope passing through the origin. This corresponds perfectly to graph (a).
• Magnetic Susceptibility ($$\chi$$) vs. Temperature ($$T$$): Diamagnetism is an intrinsic orbital property of atomic electrons and is unaffected by thermal agitation. Therefore, $$\chi$$ is independent of temperature, which is represented by a horizontal line below the temperature axis. This corresponds perfectly to graph (c).

Conclusion

The solution requires no correction. The combination representing a diamagnetic material is indeed (a) and (c), which matches Option D 

The relation between apparent dip $$\delta'$$, true dip $$\delta$$, and the angle $$\phi$$ from the magnetic meridian is $$\tan\delta' = \frac{\tan\delta}{\cos\phi}$$.

Here $$\phi = 30^\circ$$ and $$\delta' = 45^\circ$$, so $$\tan 45^\circ = \frac{\tan\delta}{\cos 30^\circ}$$.

This gives $$1 = \frac{\tan\delta}{\sqrt{3}/2}$$, so $$\tan\delta = \frac{\sqrt{3}}{2}$$.

Therefore the true dip is $$\delta = \tan^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)$$.

Upon reviewing the mathematical syntax in the previous derivation, a formatting error was found in the "Inside the Wire" bullet point where the enclosed current fraction failed to render properly. Here is the fully corrected, clean solution:

1. Derive the Magnetic Field Equations using Ampere's Law

For a long straight wire of radius $$R$$ carrying a uniformly distributed current $$I$$:

• Inside the Wire ($$r \le R$$):
  The current enclosed by an Amperian loop of radius $$r$$ is proportional to its cross-sectional area:

  $$I_{\text{enc}} = I \left( \frac{\pi r^2}{\pi R^2} \right) = I \frac{r^2}{R^2}$$

  Applying Ampere's Law ($$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$$):

  $$B_{\text{in}} \cdot 2\pi r = \mu_0 I \left(\frac{r^2}{R^2}\right) \implies B_{\text{in}} = \left(\frac{\mu_0 I}{2\pi R^2}\right) r$$

  This shows that inside the wire, the magnetic field increases linearly with radius ($$B \propto r$$).
• Outside the Wire ($$r > R$$):
  The entire current is enclosed ($I_{\text{enc}} = I$). Applying Ampere's Law:

  $$B_{\text{out}} \cdot 2\pi r = \mu_0 I \implies B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$$

  This shows that outside the wire, the magnetic field decreases hyperbolically ($$B \propto \frac{1}{r}$$).
• At the Surface ($$r = R$$):
  The magnetic field reaches its maximum value:

  $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R}$$

  Since $$B_{\text{max}}$$ is inversely proportional to the wire's radius ($$B_{\text{max}} \propto \frac{1}{R}$$), the thinner wire ($$a$$) will reach a higher peak magnetic field than the thicker wire ($$b$$) because $$a < b$$.

2. Analyze the Graph Options

Let's compile the features our graph must show:

1. From $$r = 0$$ up to the respective surfaces, both lines must be straight lines starting from the origin ($$B = 0$$ at $$r = 0$$).
2. Since $$B_{\text{max}} \propto \frac{1}{R}$$, the linear peak for wire $$a$$ must be higher than the linear peak for wire $$b$$.
3. Beyond their respective surfaces ($$r > a$$ and $$r > b$$), both curves must decay non-linearly following a hyperbolic curve ($$\propto \frac{1}{r}$$). Outside both wires ($$r > b$$), the two curves merge because the total current $$I$$ is identical.

Reviewing the choices

• Option A: Correctly shows the peak for wire $$a$$ occurring earlier and higher than the peak for wire $$b$$, with both curves sloping linearly from the origin and merging asymptotically in the outer region.

Conclusion

The correct graphical representation of the magnetic field variation is given by Option A.

For a charged particle that enters a uniform magnetic field $$\vec B$$ normally (that is, with its velocity perpendicular to the field), the magnetic force supplies the centripetal force required for uniform circular motion. First we write the force equation:

Magnetic force: $$F_{\text{mag}} = q\,v\,B$$

Centripetal force: $$F_{\text{cent}} = \dfrac{m v^{2}}{r}$$

Equating the two forces because the magnetic force itself acts as the centripetal force, we have

$$q\,v\,B = \dfrac{m v^{2}}{r}$$

Now we solve this equation for the radius $$r$$ of the circular path:

$$q\,v\,B = \dfrac{m v^{2}}{r} \;\;\Longrightarrow\;\; q\,v\,B\,r = m v^{2} \;\;\Longrightarrow\;\; r = \dfrac{m v}{q B}$$

So the radius is given by the formula $$r = \dfrac{m v}{q B}.$$

The problem involves two ions. We denote their physical quantities with subscripts 1 and 2:

Same mass: $$m_{1} = m_{2} = m$$

Charge ratio: $$q_{1} : q_{2} = 1 : 2 \;\;\Longrightarrow\;\; q_{1} = q,\; q_{2} = 2q$$

Speed ratio: $$v_{1} : v_{2} = 2 : 3 \;\;\Longrightarrow\;\; v_{1} = 2v,\; v_{2} = 3v$$

Both ions move in the same magnetic field $$B$$, so $$B$$ is common.

Using the formula for each radius, we write

$$r_{1} = \dfrac{m v_{1}}{q_{1} B} = \dfrac{m (2v)}{q B} = \dfrac{2 m v}{q B}$$

$$r_{2} = \dfrac{m v_{2}}{q_{2} B} = \dfrac{m (3v)}{2q B} = \dfrac{3 m v}{2 q B}$$

To find the required ratio of the radii, we divide $$r_{1}$$ by $$r_{2}$$:

$$\dfrac{r_{1}}{r_{2}} = \dfrac{\dfrac{2 m v}{q B}}{\dfrac{3 m v}{2 q B}}$$

Because $$m$$, $$v$$, $$q$$, and $$B$$ are common factors, we can cancel them step by step:

$$\dfrac{r_{1}}{r_{2}} = \dfrac{2}{1}\;\cdot\;\dfrac{1}{1}\;\cdot\;\dfrac{1}{1}\;\cdot\;\dfrac{2}{3} = \dfrac{4}{3}$$

Hence, $$r_{1} : r_{2} = 4 : 3.$$

Checking the options, $$4 : 3$$ corresponds to Option B.

Hence, the correct answer is Option B.

The relative permeability of a material is $$\mu_r = 1 + \chi_m$$, where $$\chi_m$$ is the magnetic susceptibility.

Given $$\chi_m = 499$$, we have $$\mu_r = 1 + 499 = 500$$.

The absolute permeability is $$\mu = \mu_r \mu_0 = 500 \times 4\pi \times 10^{-7}$$ Hm$$^{-1}$$.

Computing: $$\mu = 500 \times 4\pi \times 10^{-7} = 2000\pi \times 10^{-7} = 2\pi \times 10^{-4}$$ Hm$$^{-1}$$.

We need to determine the correct expression for the total magnetic field at point $$P(x, y)$$ due to two semi-infinite straight conductors meeting at the origin $$O(0,0)$$ at right angles.

1. Calculate the Component Fields

From the layout shown , both wires carry an identical current $$I$$ and start from the origin, extending along the positive axes:

• Wire along the y-axis: It extends from $$(0,0)$$ to $$+\infty$$. The perpendicular distance to point $$P(x, y)$$ is $$x$$. By the right-hand rule, its magnetic field points into the page ($$-\hat{k}$$):

  $$B_y = \frac{\mu_0 I}{4\pi x} \left(1 - \frac{y}{\sqrt{x^2 + y^2}}\right)$$

• Wire along the x-axis: It extends from $$(0,0)$$ to $$+\infty$$. The perpendicular distance to point $$P(x, y)$$ is $$y$$. By the right-hand rule, its magnetic field points into of the page ($$+\hat{k}$$):

  $$B_x = \frac{\mu_0 I}{4\pi y} \left(1 - \frac{x}{\sqrt{x^2 + y^2}}\right)$$

2. Compute the Net Magnetic Field ($$B_{\text{net}}$$)

$$B_{\text{net}} = B_y + B_x = \frac{\mu_0 I}{4\pi x}\left(1 - \frac{y}{\sqrt{x^2 + y^2}}\right) + \frac{\mu_0 I}{4\pi y}\left(1 - \frac{x}{\sqrt{x^2 + y^2}}\right)$$

Factoring out the common constant $$\frac{\mu_0 I}{4\pi}$$ and combining the fractions over a common denominator ($$xy\sqrt{x^2 + y^2}$$) yields:

$$B_{\text{net}} = \frac{\mu_0 I}{4\pi xy\sqrt{x^2 + y^2}} \left[ y\sqrt{x^2 + y^2} - y^2 - x\sqrt{x^2 + y^2} + x^2 \right]$$

$$B_{\text{net}} = \frac{\mu_0 I}{4\pi xy\sqrt{x^2 + y^2}} \left[ (x^2 - y^2) + (x - y)\sqrt{x^2 + y^2} \right]$$

 Option C.

We are given a magnetic needle that executes small angular oscillations in a uniform magnetic field. The standard result for such a magnet is that the time period $$T$$ of one complete oscillation is related to its moment of inertia $$I$$, its magnetic moment $$M$$, and the magnetic-field magnitude $$B$$ by

$$T \;=\; 2\pi\sqrt{\dfrac{I}{M\,B}}$$

This follows from the equation of a physical torsional pendulum, where the restoring torque is $$M B \sin\theta \approx M B\,\theta$$ for small angles, giving simple harmonic motion.

The magnet makes 10 complete oscillations in 5 s, so the time period of one oscillation is

$$T \;=\; \dfrac{\text{total time}}{\text{number of oscillations}} \;=\; \dfrac{5\ \text{s}}{10} \;=\; 0.5\ \text{s}.$$

The numerical data are

$$I = 5\times10^{-6}\ \text{kg m}^2, \qquad M = 9.85\times10^{-2}\ \text{A m}^2.$$

First we isolate $$B$$ from the period formula. Beginning with

$$T = 2\pi\sqrt{\dfrac{I}{M B}},$$

divide by $$2\pi$$:

$$\dfrac{T}{2\pi} = \sqrt{\dfrac{I}{M B}}.$$

Now square both sides:

$$\left(\dfrac{T}{2\pi}\right)^2 = \dfrac{I}{M B}.$$

Cross-multiplying gives an explicit expression for $$B$$:

$$B = \dfrac{I}{M}\;\dfrac{4\pi^{2}}{T^{2}}.$$

The question supplies the approximation $$\pi^{2}=9.85$$, so

$$4\pi^{2} = 4 \times 9.85 = 39.4.$$

Calculate the numerator:

$$I \;(4\pi^{2}) = \bigl(5\times10^{-6}\bigr)\times39.4 = 197\times10^{-6} = 1.97\times10^{-4}.$$

Calculate the denominator $$M\,T^{2}$$ step by step:

$$T^{2} = (0.5)^{2} = 0.25,$$

$$M\,T^{2} = \bigl(9.85\times10^{-2}\bigr)\times0.25 = 2.4625\times10^{-2} = 0.024625.$$

Substituting into the expression for $$B$$:

$$B = \dfrac{1.97\times10^{-4}}{0.024625} = \dfrac{1.97}{2.4625}\times10^{-2}\ \text{T}.$$

The ratio $$\dfrac{1.97}{2.4625}$$ is exactly 0.8, so

$$B = 0.8\times10^{-2}\ \text{T} = 0.008\ \text{T}.$$

Because $$1\ \text{T}=10^{3}\ \text{mT}$$, converting to millitesla gives

$$B = 0.008 \times 10^{3}\ \text{mT} = 8\ \text{mT}.$$

So, the answer is $$8\text{ mT}$$.

We have a long solenoid that originally contains a core material whose relative permeability is given as $$\mu_{r1}=500$$. The number of turns per unit length of the solenoid is $$n=1000\;\text{turns m}^{-1}$$ and the exciting current is $$I=0.75\;\text{A}$$. Later, the core is replaced by another material of the same volume but with a different relative permeability $$\mu_{r2}=750$$. We have to find the fractional change in the magnetic moment of the core.

First, recall the magnetic field strength inside a long solenoid. The standard result is

$$H = nI,$$

where $$H$$ is the magnetic field intensity, $$n$$ is the number of turns per unit length and $$I$$ is the current. Because both $$n$$ and $$I$$ remain the same before and after the replacement of the core, the quantity $$H$$ remains unchanged throughout this problem.

Next, recall the relationship between magnetisation $$M$$, magnetic susceptibility $$\chi_m$$ and the field strength $$H$$. The definition is

$$M = \chi_m\,H.$$

Magnetic susceptibility and relative permeability are connected through the familiar formula

$$\chi_m = \mu_r - 1.$$

Hence, for the first core material we have

$$\chi_{m1} = \mu_{r1} - 1 = 500 - 1 = 499,$$

and for the second core material we have

$$\chi_{m2} = \mu_{r2} - 1 = 750 - 1 = 749.$$

The magnetisations therefore become

$$M_1 = \chi_{m1}H = 499\,H,$$

$$M_2 = \chi_{m2}H = 749\,H.$$

The magnetic moment $$\mu$$ of a uniformly magnetised body is given by the product of its magnetisation $$M$$ and its volume $$V$$:

$$\mu = M\,V.$$

Because the volume of the core is stated to remain the same (namely $$V = 10^3\;\text{cm}^3$$, which we need not convert because it will cancel out), we may write

$$\mu_1 = M_1V = 499\,HV,$$

$$\mu_2 = M_2V = 749\,HV.$$

Now we compute the fractional change in the magnetic moment. The fractional change is the ratio of the change to the original value:

$$\text{Fractional change} = \frac{\mu_2 - \mu_1}{\mu_1}.$$ Substituting the expressions for $$\mu_2$$ and $$\mu_1$$ we get

$$\frac{749\,HV - 499\,HV}{499\,HV} = \frac{(749 - 499)\,HV}{499\,HV} = \frac{250}{499}.$$ Hence the fractional change can be written as $$\dfrac{x}{499}$$ with

$$x = 250.$$ Therefore, the required value is $$x = 250.$$

Hence, the correct answer is Option 250.

We have a cyclotron in which the proton is accelerated every time it passes through the gap between the two dees. The radio-frequency oscillator provides a maximum potential difference of 12 kV across the gap. Whenever the signs of the dees are opposite, the proton experiences an accelerating potential of this magnitude, gains kinetic energy, enters the circular dee, describes a semicircle, and then emerges at the gap again, where the potential has reversed so that it is once more accelerated. Hence, in one complete revolution the proton crosses the gap twice and is accelerated twice.

The energy gained in one single crossing is given by the elementary relation

$$\Delta E_{\text{cross}} = eV,$$

where $$e = 1.6 \times 10^{-19}\ \text{C}$$ is the charge on the proton and $$V = 12\ \text{kV} = 12 \times 10^{3}\ \text{V}.$$ Substituting the numbers,

$$\Delta E_{\text{cross}} = \bigl(1.6 \times 10^{-19}\bigr)\bigl(12 \times 10^{3}\bigr) = 1.6 \times 12 \times 10^{-19+3} = 19.2 \times 10^{-16}\ \text{J} = 1.92 \times 10^{-15}\ \text{J}.$$

Because the proton crosses the gap twice in one full revolution, the energy gained per revolution is

$$\Delta E_{\text{rev}} = 2 \,\Delta E_{\text{cross}} = 2 \times 1.92 \times 10^{-15}\ \text{J} = 3.84 \times 10^{-15}\ \text{J}.$$

The target speed of the proton is one sixth of the speed of light, i.e.

$$v = \frac{1}{6}\,c = \frac{1}{6}\,\bigl(3 \times 10^{8}\bigr)\ \text{m s}^{-1} = 5.0 \times 10^{7}\ \text{m s}^{-1}.$$

For this (non-relativistic) speed, the kinetic energy acquired is obtained from the classical formula

$$K = \frac{1}{2}mv^{2},$$

where $$m = 1.67 \times 10^{-27}\ \text{kg}$$ is the proton’s mass. Substituting,

$$K = \frac{1}{2}\,\bigl(1.67 \times 10^{-27}\bigr)\,\bigl(5.0 \times 10^{7}\bigr)^{2}.$$

First evaluate the square of the speed:

$$\bigl(5.0 \times 10^{7}\bigr)^{2} = 25 \times 10^{14} = 2.5 \times 10^{15}.$$

Now multiply:

$$K = \frac{1}{2} \times 1.67 \times 10^{-27} \times 2.5 \times 10^{15} = 0.835 \times 10^{-27} \times 2.5 \times 10^{15} = 2.0875 \times 10^{-12}\ \text{J}.$$

This is the total kinetic energy the proton must gain. Each revolution supplies $$\Delta E_{\text{rev}} = 3.84 \times 10^{-15}\ \text{J}.$$ Therefore the number of revolutions required is

$$N = \frac{K}{\Delta E_{\text{rev}}} = \frac{2.0875 \times 10^{-12}}{3.84 \times 10^{-15}} = \frac{2.0875}{3.84} \times 10^{3} \approx 0.5435 \times 10^{3} \approx 5.435 \times 10^{2}.$$

Keeping only whole revolutions, we need approximately 543 revolutions.

So, the answer is $$543$$.

We have an equilateral-triangle coil carrying a current in a uniform magnetic field. For a current-carrying loop, the magnitude of the magnetic torque is given by the formula

$$\tau = N\,I\,A\,B\,\sin\phi,$$

where $$N$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of the coil, $$B$$ is the magnetic-field magnitude, and $$\phi$$ is the angle between the field $$\vec B$$ and the normal to the plane of the coil.

Only one turn is mentioned, so we set $$N = 1.$$

The coil is an equilateral triangle of side $$a = 10\text{ cm} = 0.1\text{ m}.$$ The area of an equilateral triangle is

$$A = \frac{\sqrt{3}}{4}\,a^{2}.$$

Substituting $$a = 0.1\text{ m}$$, we get

$$A = \frac{\sqrt{3}}{4}\,(0.1)^{2} = \frac{\sqrt{3}}{4}\,(0.01) = 0.0025\,\sqrt{3}\;\text{m}^{2}.$$

The magnetic field is horizontal with magnitude $$B = 20\text{ mT} = 20\times10^{-3}\text{ T} = 2\times10^{-2}\text{ T}.$$

The current in the coil is $$I = 0.2\text{ A}.$$

The question states that the plane of the coil becomes parallel to the magnetic field. When the plane of the coil is parallel to $$\vec B$$, the normal to the plane is perpendicular to $$\vec B$$, giving $$\phi = 90^{\circ}$$ and hence $$\sin\phi = 1.$$

Now we substitute every quantity into the torque formula:

$$\tau = (1)\,(0.2)\,(0.0025\sqrt{3})\,(2\times10^{-2})\,(1).$$

First multiply $$0.2$$ and $$0.0025$$:

$$0.2 \times 0.0025 = 0.0005.$$

Next multiply this by $$2\times10^{-2}$$:

$$0.0005 \times 2\times10^{-2} = (0.0005 \times 2)\times10^{-2} = 0.001\times10^{-2} = 1\times10^{-5}.$$

Finally, attach the factor $$\sqrt{3}$$:

$$\tau = \sqrt{3} \times 10^{-5}\;\text{N\,m}.$$

The problem states the torque in the form $$\sqrt{x}\times10^{-5}\text{ N\,m}$$. Comparing coefficients, we have

$$x = 3.$$

So, the answer is $$3.$$

We recall the definition of magnetic moment of a current-carrying planar coil. For any coil, the magnetic moment $$m$$ is given by the formula

$$m = N\,I\,A,$$

where $$N$$ is the number of turns, $$I$$ is the current flowing through the wire and $$A$$ is the area enclosed by a single turn of the coil. This relation will be applied successively to the triangular and the square coils made from the same wire.

The total length of the conducting wire is specified to be $$24a$$. Its resistance is given as $$R$$, so when it is connected to the same voltage source of emf $$V_0$$ in either shape, the current through the wire will be identical in the two cases. Using Ohm’s law, we have

$$I = \dfrac{V_0}{R}.$$

Because $$I$$ is the same for both coils, the ratio of their magnetic moments will depend only on the respective products $$N\,A$$. Let us therefore determine $$N$$ and $$A$$ for each geometry separately.

Triangular coil  The shape is an equilateral triangle with side length $$a$$. The perimeter of one turn is $$3a$$, so the number of turns that can be formed from a total length $$24a$$ is

$$N_{\triangle}= \dfrac{24a}{3a}=8.$$

The area of a single equilateral triangle of side $$a$$ is obtained from the standard formula

$$A_{\triangle}= \dfrac{\sqrt{3}}{4}\,a^{2}.$$

Square coil  The square has side length $$a$$. Its perimeter is $$4a$$, therefore the number of turns obtained from the same wire is

$$N_{\square}= \dfrac{24a}{4a}=6.$$

The area of a square of side $$a$$ is simply

$$A_{\square}= a^{2}.$$

Ratio of magnetic moments  Using $$m=NIA$$ for each coil and canceling the common factor $$I$$, we write

$$\dfrac{m_{\triangle}}{m_{\square}} \;=\;\dfrac{N_{\triangle} A_{\triangle}} {N_{\square} A_{\square}}.$$ Substituting the numerical values just obtained, we have

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{\,8 \times \dfrac{\sqrt{3}}{4}a^{2}\,} {\,6 \times a^{2}\,}.$$

Simplifying step by step,

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{8}{4}\;\times\;\dfrac{\sqrt{3}}{6} = 2 \times \dfrac{\sqrt{3}}{6} = \dfrac{\sqrt{3}}{3}.$$

The statement of the problem expresses this same ratio in the form

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{1}{\sqrt{y}}.$$

Accordingly, we equate

$$\dfrac{1}{\sqrt{y}} = \dfrac{\sqrt{3}}{3}.$$

Cross-multiplying,

$$\sqrt{y} = \dfrac{3}{\sqrt{3}} = \sqrt{3}.$$

Finally, squaring both sides gives

$$y = 3.$$

So, the answer is $$3$$.

First recall the expression for the magnetic field in the core of a toroid. In an empty (air-filled) toroid the magnetic permeability is simply $$\mu_0$$, so the field is

$$B_0 = \mu_0\,nI,$$

where $$n = \dfrac{N}{\ell}$$ represents the number of turns per unit length and $$I$$ is the current.

When we fill the same space with a magnetic material of susceptibility $$\chi$$, its relative permeability becomes

$$\mu_r = 1 + \chi.$$

The actual permeability inside the material is therefore

$$\mu = \mu_0\,\mu_r = \mu_0(1+\chi).$$

Substituting this new permeability into the toroid formula, the magnetic field now becomes

$$B = \mu\,nI = \mu_0(1+\chi)\,nI.$$

The increase in the field strength is

$$\Delta B = B - B_0 = \mu_0(1+\chi)\,nI - \mu_0\,nI = \mu_0 nI\,\chi.$$

The fractional (or relative) increase is then

$$\frac{\Delta B}{B_0} = \frac{\mu_0 nI\,\chi}{\mu_0 nI} = \chi.$$

To turn this into a percentage increase, we multiply by 100 %:

$$\text{Percentage increase} = \chi \times 100\%.$$

For aluminium we are given

$$\chi = 2.2 \times 10^{-5}.$$

Hence

$$\text{Percentage increase} = 2.2 \times 10^{-5} \times 100 = 2.2 \times 10^{-3}.$$

According to the problem statement, this percentage is expressed as $$\dfrac{x}{10^{4}}$$. Setting the two expressions equal gives

$$\frac{x}{10^{4}} = 2.2 \times 10^{-3}.$$

Now we solve for $$x$$:

$$x = 2.2 \times 10^{-3} \times 10^{4} = 2.2 \times 10^{1} = 22.$$

So, the answer is $$22$$.

We need to determine the torque experienced by the magnetic dipole $$m_2$$ due to the magnetic field produced by the magnetic dipole $$m_1$$.

1. Identify the Given Parameters

From the problem parameters , we have two short magnetic dipoles with the following values:

• Magnetic moment of the first dipole ($$m_1$$) = $$1\text{ A m}^2$$
• Magnetic moment of the second dipole ($$m_2$$) = $$1\text{ A m}^2$$
• Distance between the points O and P ($$r$$) = $$1\text{ m}$$

Based on standard dipole configurations for calculating torque, the dipole $$m_1$$ at point O creates an axial magnetic field at the position of dipole $$m_2$$ at point P, where $$m_2$$ is oriented perpendicular to this field.

2. Calculate the Magnetic Field ($$B_1$$) Due to $$m_1$$

The magnetic field produced by a short magnetic dipole at a point on its axial line at distance $$r$$ is given by the formula:

$$B_1 = \frac{\mu_0}{4\pi} \frac{2m_1}{r^3}$$

Substituting the standard value of $$\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m A}^{-1}$$ and the given problem constants:

$$B_1 = 10^{-7} \times \frac{2 \times 1}{(1)^3} = 2 \times 10^{-7}\text{ T}$$

3. Determine the Torque experienced by $$m_2$$

The torque ($$\tau$$) acting on a magnetic dipole placed in a uniform external magnetic field is given by the vector cross product formula:

$$\tau = m_2 \cdot B_1 \cdot \sin(\theta)$$

Since the dipole $$m_2$$ is oriented perpendicular ($$\theta = 90^\circ$$) to the axial magnetic field line produced by $$m_1$$, the value of $$\sin(90^\circ) = 1$$. Substituting these values gives:

$$\tau = 1 \times (2 \times 10^{-7}) \times 1 = 2 \times 10^{-7}\text{ N m}$$

4. Match with the Requested Format

The problem asks for the torque in the expression format of $$\text{Value} \times 10^{-7}\text{ N m}$$:

$$\tau = 2 \times 10^{-7}\text{ N m} \implies \text{Value} = 2$$

Conclusion

The value of the torque experienced by the magnetic dipole is 2.

We have a particle of mass $$m$$ and charge $$q$$. Its initial velocity is purely along the $$y$$-axis, so we may write it as $$\vec v(0)=v_0\,\hat j$$. An electric field $$\vec E$$ and a magnetic field $$\vec B$$ are both directed along the $$x$$-axis; in symbols we write $$\vec E=E_0\,\hat i$$ and $$\vec B=B_0\,\hat i$$.

The force acting on a charged particle in simultaneous electric and magnetic fields is given by the Lorentz formula

$$\vec F = q\,(\vec E + \vec v \times \vec B).$$

Dividing by the mass, we obtain the acceleration

$$\frac{d\vec v}{dt}= \frac{q}{m}\,\bigl(\vec E + \vec v \times \vec B\bigr).$$

Let us resolve every quantity into Cartesian components. We denote the velocity as $$\vec v=(v_x,\,v_y,\,v_z).$$ The electric field has only an $$x$$-component $$E_0$$, while the magnetic field has only an $$x$$-component $$B_0$$. Therefore

$$\vec E=(E_0,\,0,\,0), \qquad \vec B=(B_0,\,0,\,0).$$

The magnetic part of the force requires the cross-product $$\vec v \times \vec B$$. Writing out the determinant

$$ \vec v \times \vec B= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] v_x & v_y & v_z\\[2pt] B_0 & 0 & 0 \end{vmatrix} = \hat i\,(v_y\cdot 0-v_z\cdot 0)\;-\; \hat j\,(v_x\cdot 0-v_z\cdot B_0)\;+\; \hat k\,(v_x\cdot 0-v_y\cdot B_0). $$

Simplifying term by term we get

$$\vec v \times \vec B =(0,\;v_z B_0,\;-v_y B_0).$$

Substituting the electric and magnetic contributions into the acceleration equation gives the three component equations

$$m\,\frac{dv_x}{dt}=qE_0,$$

$$m\,\frac{dv_y}{dt}=q\,v_z B_0,$$

$$m\,\frac{dv_z}{dt}=-q\,v_y B_0.$$

We now analyse each direction.

1. Along the $$x$$-axis the acceleration is constant because the right-hand side is the constant $$qE_0$$. Hence

$$\frac{dv_x}{dt}= \frac{qE_0}{m}\equiv a,$$

where we have introduced $$a=\dfrac{qE_0}{m}$$ for convenience. Integrating with the initial condition $$v_x(0)=0$$ (because the initial velocity had no $$x$$-component) we get

$$v_x(t)=a\,t=\frac{qE_0}{m}\,t.$$

2. The $$y$$- and $$z$$-components are coupled. It is helpful to introduce the cyclotron frequency

$$\omega=\frac{qB_0}{m}.$$

The two differential equations now read

$$\frac{dv_y}{dt}= \omega\,v_z,$$

$$\frac{dv_z}{dt}= -\omega\,v_y.$$

Differentiate the first of these once more with respect to time:

$$\frac{d^2 v_y}{dt^2}= \omega\,\frac{dv_z}{dt}=\omega\,(-\omega v_y)=-\omega^2\,v_y.$$

This is the standard simple-harmonic-motion equation $$\dfrac{d^2y}{dt^2}=-\omega^2 y$$ whose solution is a sinusoid. Using the initial conditions $$v_y(0)=v_0$$ and $$v_z(0)=0$$ we obtain

$$v_y(t)=v_0\cos(\omega t),$$

and, from the relation $$\dfrac{dv_y}{dt}=\omega v_z,$$

$$v_z(t)= -\,v_0\sin(\omega t).$$

We now compute the square of the speed at any instant. The speed is the magnitude of $$\vec v$$, so

$$ \begin{aligned} v^2(t)&=v_x^2(t)+v_y^2(t)+v_z^2(t)\\[4pt] &=\bigl(a t\bigr)^2+\bigl(v_0\cos\omega t\bigr)^2+\bigl(-v_0\sin\omega t\bigr)^2\\[4pt] &=a^2 t^2+v_0^2\bigl(\cos^2\omega t+\sin^2\omega t\bigr). \end{aligned} $$

Because $$\cos^2\theta+\sin^2\theta=1$$, the trigonometric terms combine neatly to unity, leaving

$$v^2(t)=v_0^2 + a^2 t^2.$$

We are asked to find the time at which the speed doubles, i.e. when $$v(t)=2v_0$$. Squaring this requirement gives

$$\bigl(2v_0\bigr)^2 = v_0^2 + a^2 t^2.$$

Simplifying step by step,

$$4v_0^2 = v_0^2 + a^2 t^2,$$

$$4v_0^2 - v_0^2 = a^2 t^2,$$

$$3v_0^2 = a^2 t^2,$$

$$t^2 = \frac{3v_0^2}{a^2}.$$

Taking the positive square root (time is positive) we have

$$t = \frac{\sqrt{3}\,v_0}{a}.$$

Finally substitute back $$a=\dfrac{qE_0}{m}$$:

$$t = \frac{\sqrt{3}\,v_0}{\dfrac{qE_0}{m}} =\frac{\sqrt{3}\,m\,v_0}{qE_0}.$$

This time corresponds exactly to Option C in the given list.

Hence, the correct answer is Option C.

We first write the Lorentz-force law for a charged particle moving in a magnetic field:

$$\vec F = q\,\vec v \times \vec B.$$

The magnetic field is along the positive $$z$$-axis,

$$\vec B = B_0 \hat k,$$

while the particle is shot towards the screen with velocity

$$\vec v = -\,v\hat i \qquad (v \neq 0).$$

Taking the cross-product we get

$$\vec v \times \vec B \;=\; (-\,v\hat i)\times(B_0\hat k) = -\,vB_0\;(\hat i \times \hat k) = -\,vB_0\,(-\hat j) = +\,vB_0\,\hat j,$$

so

$$\vec F = q\,vB_0\,\hat j.$$

The force (and therefore the acceleration) is along $$+y$$, perpendicular to the initial velocity. A constant force perpendicular to the velocity produces uniform circular motion in the plane perpendicular to $$\vec B$$ (the $$x\!-\!y$$ plane).

For such motion the magnetic force supplies the necessary centripetal force. Stating the formula,

$$\text{Centripetal force} = \frac{mv^{2}}{r}, \qquad \text{Magnetic force} = qvB_0,$$

and equating them gives

$$qvB_0 = \frac{mv^{2}}{r}.$$

Solving for the radius $$r$$ of the circular path,

$$r = \frac{mv}{qB_0}.$$

Let the screen be the plane $$x = 0$$. The particle starts from the point $$P(d,0)$$ and its initial velocity is towards $$-x$$. Because the acceleration is along $$+y$$, the centre of the circle lies a distance $$r$$ above the starting point, i.e. at $$C(d,\,r)$$. Hence the circle has centre abscissa $$x = d$$ and radius $$r$$. The left-most point of this circle (the point of smallest $$x$$) is therefore at

$$x_{\min} = d - r.$$

If the particle is to miss the screen, this left-most point must stay to the right of the screen, i.e.

$$x_{\min} > 0 \;\;\Longrightarrow\;\; d - r > 0 \;\;\Longrightarrow\;\; r < d.$$

The borderline (minimum) speed that just manages to avoid the screen occurs when the trajectory merely grazes it, i.e. when

$$r = d.$$

Substituting $$r = d$$ in the expression $$r = \dfrac{mv}{qB_0}$$, we obtain

$$d = \frac{mv_{\min}}{qB_0} \;\;\Longrightarrow\;\; v_{\min} = \frac{qB_0\,d}{m}.$$

Thus the least speed required so that the particle never reaches the screen is

$$v_{\min} = \frac{q\,d\,B_0}{m}.$$

Comparing with the given options, this corresponds to Option C.

Hence, the correct answer is Option C.

We have a thin circular loop of radius $$a$$, carrying a current $$I$$ and possessing mass $$m$$. In a uniform magnetic field $$B$$ directed perpendicular to the plane of the loop, the loop behaves like a magnetic dipole. Its magnetic dipole moment is, by definition,

$$\mu = I \,A = I\,(\pi a^{2}).$$

Next, we need the moment of inertia of the loop about the diameter about which it is allowed to oscillate. For a thin circular ring, the perpendicular-axis theorem gives

$$I_{x}+I_{y}=I_{z}.$$

Here $$I_{z}=m a^{2}$$ (axis through the centre and perpendicular to the plane). Because the two diameters in the plane are equivalent, $$I_{x}=I_{y}$$; hence

$$2I_{x}=m a^{2}\quad\Longrightarrow\quad I_{x}=I_{y}=\frac{1}{2}m a^{2}.$$

Thus the moment of inertia about the diameter of oscillation is

$$I_{\text{m.o.i.}}=\frac{1}{2}m a^{2}.$$

When the loop is given a small angular displacement $$\theta$$ from its equilibrium position, the restoring magnetic torque is

$$\tau = \mu B \sin\theta.$$

For small angles, $$\sin\theta \approx \theta$$, so

$$\tau \approx \mu B\,\theta.$$

The rotational form of Newton’s second law states

$$I_{\text{m.o.i.}}\;\frac{d^{2}\theta}{dt^{2}} = -\tau,$$

the negative sign indicating that the torque is restoring. Substituting the expressions for $$\tau$$ and $$I_{\text{m.o.i.}}$$, we obtain

$$\frac{1}{2}m a^{2}\;\frac{d^{2}\theta}{dt^{2}} \;=\; -\,\mu B\,\theta.$$

Simplifying,

$$\frac{d^{2}\theta}{dt^{2}} + \underbrace{\left(\frac{2\mu B}{m a^{2}}\right)}_{\omega^{2}}\theta = 0.$$

This is the standard differential equation of simple harmonic motion, $$\displaystyle \frac{d^{2}\theta}{dt^{2}}+\omega^{2}\theta = 0,$$ whose angular frequency is

$$\omega = \sqrt{\frac{2\mu B}{m a^{2}}}.$$

The time period is related to angular frequency by the formula

$$T = \frac{2\pi}{\omega}.$$

Substituting $$\omega$$,

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\mu B}}.$$

Now we replace $$\mu$$ by $$I\pi a^{2}$$:

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\,I\pi a^{2} B}} = 2\pi\sqrt{\frac{m}{2I\pi B}}.$$

Notice that $$a^{2}$$ cancels out. Extracting the common factors inside the square root,

$$T = 2\pi\;\frac{\sqrt{m}}{\sqrt{2I\pi B}} = \frac{2\sqrt{\pi}\,\sqrt{m}}{\sqrt{2I B}} = \sqrt{\frac{2\pi m}{I B}}.$$

Hence, the correct answer is Option C.

Perpendicular distance from the centre to each side of a regular hexagon:

$$d = a \cos(30^\circ) = 0.1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{20}\ \text{m}$$

Angles subtended by each side at the centre: $$\theta_1 = \theta_2 = 30^\circ$$

Magnetic field due to one side of a single turn:

$$B_1 = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2) = \frac{\mu_0 I}{4\pi \left(\frac{\sqrt{3}}{20}\right)}(\sin 30^\circ + \sin 30^\circ)$$

$$B_1 = \frac{20\mu_0 I}{4\pi \sqrt{3}}\left(\frac{1}{2} + \frac{1}{2}\right) = \frac{5\mu_0 I}{\pi \sqrt{3}}$$

$$B = 6 \times n \times B_1 = 6 \times 50 \times \frac{5\mu_0 I}{\pi \sqrt{3}}$$

$$B = \frac{1500}{\sqrt{3}}\frac{\mu_0 I}{\pi} = 500\sqrt{3}\left(\frac{\mu_0 I}{\pi}\right)$$

Time period of one revolution: $$T = \frac{2\pi m}{qB}$$

Pitch of the helical path: $$p = v \cos\theta \cdot T = \frac{2\pi m v \cos\theta}{qB}$$

Total length traveled along the field: $$l = N \cdot p = \frac{2\pi m v N \cos\theta}{qB}$$

$$l = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^5 \times 10 \times \cos 60^\circ}{1.6 \times 10^{-19} \times 0.3}$$

$$l = \frac{2 \times 3.14 \times 1.67 \times 4 \times 10 \times 0.5 \times 10^{-22}}{0.48 \times 10^{-19}} = \frac{41.95 \times 10^{-21}}{4.8 \times 10^{-20}} \approx 0.44\ \text{m}$$

We have a stream of protons that enter a uniform magnetic field with speed $$v = 4 \times 10^{5}\,\text{m s}^{-1}$$ making an angle $$\theta = 60^{\circ}$$ with the field. Because the velocity has a component parallel to the field and another component perpendicular to it, each proton follows a helical path. The physical quantity we are asked to find is the pitch of that helix, i.e. the distance travelled parallel to the magnetic field in one complete revolution about the field lines.

First we resolve the velocity into two perpendicular components. Using $$v_{\parallel}=v\cos\theta$$ for the component parallel to the field and $$v_{\perp}=v\sin\theta$$ for the component perpendicular to the field, we write

$$v_{\parallel}=v\cos 60^{\circ}=4\times10^{5}\times\frac12=2\times10^{5}\;\text{m s}^{-1}.$$

The perpendicular component is not directly required for the pitch, but it is this component that causes the circular motion. The circular motion has an angular frequency $$\omega$$ determined by the magnetic force. For a charge $$q$$ moving in a uniform magnetic field $$B$$, the formula for the angular frequency is stated as

$$\omega=\frac{qB}{m},$$

where $$m$$ is the mass of the particle. Substituting the given values $$q=1.69\times10^{-19}\;\text{C},\; B=0.3\;\text{T},\; m=1.67\times10^{-27}\;\text{kg},$$ we obtain

$$\omega=\frac{1.69\times10^{-19}\times0.3}{1.67\times10^{-27}}.$$

Multiplying in the numerator gives

$$1.69\times0.3=0.507,$$

so

$$\omega=\frac{0.507\times10^{-19}}{1.67\times10^{-27}} =\frac{5.07\times10^{-20}}{1.67\times10^{-27}} =3.03\times10^{7}\;\text{s}^{-1}\;(\text{approximately}).$$

The time period $$T$$ of one revolution is the reciprocal of the frequency, and for angular frequency we use

$$T=\frac{2\pi}{\omega}.$$

Substituting $$\omega=3.03\times10^{7}\;\text{s}^{-1}$$ gives

$$T=\frac{2\pi}{3.03\times10^{7}} =\frac{6.283}{3.03}\times10^{-7} =2.07\times10^{-7}\;\text{s}.$$

Now the pitch $$p$$ is the distance travelled along the magnetic field in one time period:

$$p=v_{\parallel}\,T.$$

Using $$v_{\parallel}=2\times10^{5}\;\text{m s}^{-1}$$ and $$T=2.07\times10^{-7}\;\text{s},$$ we have

$$p=2\times10^{5}\times2.07\times10^{-7} =4.14\times10^{-2}\;\text{m}.$$

To express this in centimetres, we multiply by 100:

$$p=4.14\times10^{-2}\times100=4.14\;\text{cm}.$$

The closest value among the options is 4 cm.

Hence, the correct answer is Option D.

We need the magnetic force on a moving charge. The magnetic (Lorentz) force formula is stated first:

$$\vec{F}=q\,\left(\vec{v}\times\vec{B}\right).$$

Here the charge is $$q = 1\,\mu\text{C}=1\times10^{-6}\,\text{C},$$ the velocity is $$\vec{v}=2\hat i+3\hat j+4\hat k\;\text{m s}^{-1},$$ and the magnetic field is $$\vec{B}=\left(5\hat i+3\hat j-6\hat k\right)\times10^{-3}\,\text{T}.$$

We first evaluate the cross product $$\vec{v}\times\vec{B}.$$ Writing out the determinant:

$$$ \vec{v}\times\vec{B}= \begin{vmatrix} \hat i & \hat j & \hat k\\[4pt] 2 & 3 & 4\\[4pt] 5\times10^{-3} & 3\times10^{-3} & -6\times10^{-3} \end{vmatrix}. $$$

Because every component of $$\vec{B}$$ already contains the common factor $$10^{-3},$$ we keep that factor outside for clarity and perform the determinant with the numerical parts $$5,\,3,\, -6$$ only. Thus,

$$$ \vec{v}\times\vec{B}=10^{-3}\! \begin{vmatrix} \hat i & \hat j & \hat k\\[4pt] 2 & 3 & 4\\[4pt] 5 & 3 & -6 \end{vmatrix}. $$$

Expanding the 3×3 determinant step by step:

$$$ \begin{aligned} \vec{v}\times\vec{B} &=10^{-3}\Bigl[\hat i\,(3\!\cdot\!(-6)-4\!\cdot\!3)\;-\;\hat j\,(2\!\cdot\!(-6)-4\!\cdot\!5)\;+\;\hat k\,(2\!\cdot\!3-3\!\cdot\!5)\Bigr] \\[4pt] &=10^{-3}\Bigl[\hat i\,(-18-12)\;-\;\hat j\,(-12-20)\;+\;\hat k\,(6-15)\Bigr] \\[4pt] &=10^{-3}\Bigl[\hat i\,(-30)\;+\;\hat j\,(32)\;+\;\hat k\,(-9)\Bigr] \\[4pt] &=\left(-30\hat i+32\hat j-9\hat k\right)\times10^{-3}. \end{aligned} $$$

Now we multiply this result by the charge $$q=1\times10^{-6}\,\text{C}$$ to obtain the force:

$$$ \vec{F}=q\left(\vec{v}\times\vec{B}\right) =1\times10^{-6}\,\text{C}\;\times\;\left(-30\hat i+32\hat j-9\hat k\right)\times10^{-3}. $$$

Combining the powers of ten, $$10^{-6}\times10^{-3}=10^{-9},$$ so

$$$ \vec{F}=\left(-30\hat i+32\hat j-9\hat k\right)\times10^{-9}\,\text{N}. $$$

The problem statement itself writes the force in the form $$\vec{F}\times10^{-9}\,\text{N},$$ so the required vector $$\vec{F}$$ is simply

$$\vec{F}=-30\hat i+32\hat j-9\hat k.$$ Hence, the correct answer is Option B.

We are told that a bar magnet placed in a uniform magnetic field experiences a torque. For a magnetic dipole, the relation between the torque $$\tau$$, the magnetic dipole moment $$M$$, the magnetic field $$B$$ and the angle $$\theta$$ between $$\vec M$$ and $$\vec B$$ is stated first:

$$\tau \;=\; M\,B\,\sin\theta$$

The numerical values given are

$$\tau = 0.018\ \text{N m}, \qquad B = 0.06\ \text{T}, \qquad \theta = 30^\circ.$$

We substitute these values into the formula to isolate $$M$$. We begin by writing

$$M = \dfrac{\tau}{B\,\sin\theta}.$$

Now, since $$\sin 30^\circ = \dfrac12,$$ we have

$$B\,\sin\theta \;=\; 0.06 \times \dfrac12 \;=\; 0.03.$$

So

$$M = \dfrac{0.018}{0.03} = 0.6\ \text{A m}^2.$$

Next we calculate the work needed to turn the magnet from its stable to its unstable equilibrium position. The potential energy $$U$$ of a magnetic dipole in a field is

$$U = -M\,B\,\cos\theta.$$

The stable equilibrium corresponds to $$\theta = 0^\circ$$ (magnet aligned with the field), and the unstable equilibrium corresponds to $$\theta = 180^\circ$$ (magnet anti-aligned). We compute the energies at these two orientations.

At $$\theta = 0^\circ$$:

$$U_{\text{stable}} = -M\,B\,\cos 0^\circ = -M\,B \times 1 = -M\,B.$$

At $$\theta = 180^\circ$$:

$$U_{\text{unstable}} = -M\,B\,\cos 180^\circ = -M\,B \times (-1) = +M\,B.$$

The work required is the increase in potential energy while going from the stable to the unstable state:

$$W = U_{\text{unstable}} - U_{\text{stable}} = (M\,B) - (-M\,B) = 2\,M\,B.$$

Substituting the values $$M = 0.6\ \text{A m}^2$$ and $$B = 0.06\ \text{T}$$ we get

$$W = 2 \times 0.6 \times 0.06 = 2 \times 0.036 = 0.072\ \text{J}.$$

Writing this in scientific notation,

$$W = 7.2 \times 10^{-2}\ \text{J}.$$

Hence, the correct answer is Option C.

Magnetic moment of loop ABCDA in the XY-plane: $$\vec{M}_1 = I(a \times b)\hat{k} = Iab\hat{k}$$

Magnetic moment of loop ADEFA in the XZ-plane: $$\vec{M}_2 = I(a \times b)\hat{j} = Iab\hat{j}$$

Total magnetic dipole moment vector: $$\vec{M} = \vec{M}_1 + \vec{M}_2 = Iab(\hat{j} + \hat{k})$$

Magnitude of the total magnetic moment: $$M = \sqrt{(Iab)^2 + (Iab)^2} = \sqrt{2}abI$$

Direction unit vector: $$\hat{u} = \frac{\vec{M}}{M} = \frac{Iab(\hat{j} + \hat{k})}{\sqrt{2}abI} = \left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$$

$$B = \mu_0 n I$$

Since the electron is shot radially outwards from the axis, its velocity $$\vec{v}$$ is perpendicular to the axial magnetic field $$\vec{B}$$, causing it to trace a circular path that passes through the center (axis).

To ensure the electron avoids hitting the inner wall of the solenoid of radius $$R$$, the maximum diameter of its circular trajectory cannot exceed the solenoid's radius:

$$2r_{\text{max}} = R \implies r_{\text{max}} = \frac{R}{2}$$

$$\frac{m v_{\text{max}}}{e B} = \frac{R}{2}$$

$$v_{\text{max}} = \frac{e B R}{2m} = \frac{e \mu_0 n I R}{2m}$$

We have an electron whose velocity is strictly along the $$+x$$-axis, so we write

$$\vec v = 6 \times 10^{6}\,\text{m s}^{-1}\,\hat i.$$

The region also possesses a uniform electric field pointing along $$+y$$, that is

$$\vec E = 300\,\text{V cm}^{-1}\,\hat j.$$

First, we convert this electric field into SI units. Since $$1\,\text{cm}=10^{-2}\,\text{m},$$ we obtain

$$300\,\text{V cm}^{-1}=300 \times 10^{2}\,\text{V m}^{-1}=3.0 \times 10^{4}\,\text{V m}^{-1}.$$

For a charge $$q$$ moving in simultaneous electric and magnetic fields, the Lorentz force formula is stated as

$$\vec F = q\bigl(\vec E + \vec v \times \vec B\bigr).$$

We want the electron to continue moving only along the $$x$$-direction, which means the net transverse force must vanish. Hence we require

$$\vec F = \vec 0 \Longrightarrow \vec E + \vec v \times \vec B = \vec 0.$$

Re-arranging gives

$$\vec v \times \vec B = -\vec E.$$

Because the electron carries charge $$q=-e,$$ its electric force is $$\vec F_E = -e\vec E.$$ Since $$\vec E$$ is along $$+y,$$ $$\vec F_E$$ is along $$-y.$$ Therefore the magnetic force $$\vec F_B = q(\vec v \times \vec B)$$ must point along $$+y$$ in order to exactly cancel $$\vec F_E.$$ To secure that direction we inspect the cross-product:

$$\vec v = v\,\hat i, \qquad \vec B = B\,\hat k \;(\text{assume}).$$

Using the right-hand rule,

$$\hat i \times \hat k = -\hat j,$$

so

$$\vec v \times \vec B = vB(-\hat j).$$

Multiplying by the electron’s charge $$q=-e$$ gives

$$q(\vec v \times \vec B) = (-e)\bigl(vB(-\hat j)\bigr)=+e vB\,\hat j,$$

which indeed points along $$+y$$ as required. Hence our initial assumption that $$\vec B$$ is along $$+z$$ ($$\hat k$$) is correct. The magnetic field must therefore be directed along $$+z$$.

To find its magnitude, we equate the magnitudes of the electric and magnetic forces:

$$eE = e v B \quad\Longrightarrow\quad B = \frac{E}{v}.$$

Substituting $$E = 3.0 \times 10^{4}\,\text{V m}^{-1}$$ and $$v = 6 \times 10^{6}\,\text{m s}^{-1},$$ we obtain

$$B = \frac{3.0 \times 10^{4}}{6 \times 10^{6}}\,\text{T} = 0.5 \times 10^{-2}\,\text{T} = 5 \times 10^{-3}\,\text{T}.$$

Thus, a magnetic field of magnitude $$5 \times 10^{-3}\,\text{T}$$ directed along $$+z$$ will ensure the electron continues straight along the $$x$$-axis.

Hence, the correct answer is Option C.

Total magnetic flux through any closed Gaussian surface: $$\oint \vec{B} \cdot d\vec{A} = 0$$

Let a closed surface encompass the entire upper hemisphere bounded by the infinite plane:

$$\phi_{\text{total}} = \phi_{\text{coil\_area}} + \phi_{\text{remaining\_plane}} + \phi_{\text{hemisphere\_at\_infinity}} = 0$$

Since magnetic field falls off as $$B \propto \frac{1}{r^3}$$ for a dipole: $$\phi_{\text{hemisphere\_at\_infinity}} = 0$$

Flux contribution from the two regions of the infinite coplanar boundary: $$\phi_0 + \phi_i = 0 \implies \phi_i = -\phi_0$$

Retentivity is the value of $$B$$ when $$H = 0$$: $$B_{\text{retentivity}} = 1.0\text{ T}$$

Coercivity is the value of $$|H|$$ required to reduce $$B$$ to zero: $$H_{\text{coercivity}} = 50\text{ A/m}$$

Saturation is the maximum value of $$B$$ where the curve flattens out: $$B_{\text{saturation}} = 1.5\text{ T}$$

We consider a very long straight wire of radius $$a$$ that carries a steady current $$I$$. The current is said to be uniformly distributed over the circular cross-section, which means the current density $$J$$ (current per unit area) is the same at every point inside the wire.

Because the situation is cylindrically symmetric, we use Ampère’s circuital law in its integral form, which states

$$\oint \vec B \cdot d\vec l \;=\;\mu_0 I_{\text{encl}}$$

where $$I_{\text{encl}}$$ is the current enclosed by the chosen circular Amperian path and $$\mu_0$$ is the permeability of free space. For a circle of radius $$r$$ centred on the axis of the wire, $$\vec B$$ is tangential and has the same magnitude everywhere on the circle, so the left side becomes $$B\,(2\pi r)$$. Thus

$$B\,(2\pi r)=\mu_0 I_{\text{encl}} \quad\Longrightarrow\quad B=\frac{\mu_0 I_{\text{encl}}}{2\pi r}$$

Now we need the magnetic field at two different distances from the axis: (i) at $$r_1=\dfrac{a}{3}$$ (inside the wire) and (ii) at $$r_2=2a$$ (outside the wire).

1. Field at $$r_1=\dfrac{a}{3}$$ (inside the wire)

Because the point lies inside the conductor, only the current contained in the smaller circle of radius $$r_1$$ contributes. The current density is

$$J=\frac{I}{\pi a^{2}}\;,$$

so the enclosed current is

$$I_{\text{encl}}=J\;(\text{area within }r_1)=\frac{I}{\pi a^{2}}\;(\pi r_1^{2})=\frac{I r_1^{2}}{a^{2}}.$$

Substituting $$r_1=\dfrac{a}{3}$$ gives

$$I_{\text{encl}}=\frac{I}{a^{2}}\left(\frac{a}{3}\right)^{2}=\frac{I a^{2}}{9a^{2}}=\frac{I}{9}.$$

Putting this value into the expression for $$B$$, we get the magnetic field at $$r_1$$:

$$B_1=\frac{\mu_0 I_{\text{encl}}}{2\pi r_1} =\frac{\mu_0 \left(\dfrac{I}{9}\right)}{2\pi \left(\dfrac{a}{3}\right)} =\frac{\mu_0 I}{9}\;\frac{3}{2\pi a} =\frac{\mu_0 I}{6\pi a}.$$

2. Field at $$r_2=2a$$ (outside the wire)

For a point outside, the entire current $$I$$ is enclosed, i.e., $$I_{\text{encl}}=I$$. Using the same formula, we have

$$B_2=\frac{\mu_0 I}{2\pi r_2} =\frac{\mu_0 I}{2\pi (2a)} =\frac{\mu_0 I}{4\pi a}.$$

3. Required ratio $$\dfrac{B_1}{B_2}$$

We now divide $$B_1$$ by $$B_2$$:

$$\frac{B_1}{B_2} =\frac{\dfrac{\mu_0 I}{6\pi a}}{\dfrac{\mu_0 I}{4\pi a}} =\left(\frac{\mu_0 I}{6\pi a}\right)\left(\frac{4\pi a}{\mu_0 I}\right) =\frac{4}{6} =\frac{2}{3}.$$

Hence, the ratio of the magnetic fields at the two specified distances is $$\dfrac{2}{3}$$.

Hence, the correct answer is Option A.

$$I \propto \frac{B}{T} \implies \frac{I_2}{I_1} = \left(\frac{B_2}{B_1}\right) \left(\frac{T_1}{T_2}\right)$$

$$\frac{I_2}{6} = \left(\frac{0.3}{0.4}\right) \left(\frac{4}{24}\right) = \left(\frac{3}{4}\right) \left(\frac{1}{6}\right) = \frac{1}{8}$$

$$I_2 = \frac{6}{8} = 0.75\text{ A/m}$$

Magnetic permeability of a perfect diamagnet: $$\mu_r = 0$$

Magnetic field inside the bulk diamagnetic material: $$B_{\text{inside}} = \mu_r B = 0$$

Since the external magnetic field lines cannot penetrate the perfectly diamagnetic outer sphere, no magnetic flux reaches the inner cavity.

$$\text{Field inside the central cavity} = 0$$

Let the centre of the circular portion be O and its radius be $$R$$. We have to add the magnetic fields produced at O by

(i) the complete circular turn $$DMND$$ and
(ii) the two very long straight parts $$AB$$ and $$BC$$ which touch the circle tangentially at the points $$N$$ and $$D$$ respectively.

Field due to the circular turn
For a circular loop carrying current $$I$$, the magnetic field at its centre is given by the well-known result

$$B_{\text{circle}}=\frac{\mu_0 I}{2R}\;.$$

It will be directed perpendicular to the plane of the loop. Writing the same expression with a denominator $$2\pi R$$ (so that all later terms have a common factor) we have

$$B_{\text{circle}}=\frac{\mu_0 I}{2\pi R}\,\pi\;.$$

Field due to the straight part AB
For a finite straight conductor the Biot-Savart result is

$$B=\frac{\mu_0 I}{4\pi r}\,(\sin\theta_1+\sin\theta_2),$$

where $$r$$ is the perpendicular distance of the point from the wire and $$\theta_1,\;\theta_2$$ are the angles which the lines joining the point to the two ends of the conductor make with the conductor itself.

The perpendicular distance of O from AB is clearly $$R$$ (AB is a tangent). One end of AB (the point A) is at infinity, so the line AO is almost along AB itself; hence $$\theta_A=0^\circ$$ and $$\sin\theta_A=0$$. The other end is B. Because B has the coordinates $$(R,\;R)$$ while the wire AB is horizontal, the line OB makes an angle of $$45^\circ$$ with AB, giving $$\theta_B=45^\circ$$ and $$\sin\theta_B=\dfrac1{\sqrt2}$$. Substituting,

$$B_{AB}=\frac{\mu_0 I}{4\pi R}\left(0+\frac1{\sqrt2}\right)=\frac{\mu_0 I}{4\pi R\sqrt2}\;.$$

Field due to the straight part BC
Exactly the same geometry holds for BC, because BC is a vertical tangent lying a distance $$R$$ from O. Again one end (C) is at infinity giving $$\sin\theta_C=0$$, while the nearer end B subtends $$45^\circ$$ with BC giving $$\sin\theta_B=\dfrac1{\sqrt2}$$. Therefore

$$B_{BC}=\frac{\mu_0 I}{4\pi R}\left(\frac1{\sqrt2}+0\right)=\frac{\mu_0 I}{4\pi R\sqrt2}\;.$$

Directions of the straight-wire fields
If the current reaches the junction B from A (i.e.\ it flows to the left along AB) and then turns to flow upward along BC, the right-hand rule shows that both $$B_{AB}$$ and $$B_{BC}$$ point out of the plane of the paper. Hence the two contributions add:

$$B_{\text{straight}}=B_{AB}+B_{BC}=\frac{\mu_0 I}{4\pi R\sqrt2}+\frac{\mu_0 I}{4\pi R\sqrt2} =\frac{\mu_0 I}{2\pi R\sqrt2}\;.$$

Total field at the centre
Adding the circular and straight-wire fields (both are perpendicular to the plane and hence add algebraically)

$$\begin{aligned} B_{\text{net}}&=B_{\text{circle}}+B_{\text{straight}}\\[4pt] &=\frac{\mu_0 I}{2\pi R}\,\pi+\frac{\mu_0 I}{2\pi R}\,\frac1{\sqrt2}\\[4pt] &=\frac{\mu_0 I}{2\pi R}\left(\pi+\frac1{\sqrt2}\right). \end{aligned}$$

Hence, the correct answer is Option A.

$$\theta_A = 360^\circ - 90^\circ = 270^\circ = \frac{3\pi}{2}\ \text{rad}$$

$$\theta_B = 360^\circ - 60^\circ = 300^\circ = \frac{5\pi}{3}\ \text{rad}$$

$$B_A = \frac{\mu_0 I_A}{4\pi R_A} \theta_A$$

$$B_B = \frac{\mu_0 I_B}{4\pi R_B} \theta_B$$

$$\frac{B_A}{B_B} = \left(\frac{I_A}{I_B}\right) \times \left(\frac{R_B}{R_A}\right) \times \left(\frac{\theta_A}{\theta_B}\right)$$

$$\frac{B_A}{B_B} = \left(\frac{2}{3}\right) \times \left(\frac{4}{2}\right) \times \left(\frac{270^\circ}{300^\circ}\right) = \left(\frac{2}{3}\right) \times 2 \times \left(\frac{9}{10}\right) = \frac{36}{30} = \frac{6}{5}$$

We are given that a proton possesses a kinetic energy of $$1\ \text{MeV}$$ and is travelling from south to north. First, we convert this energy into joules because the usual mechanical formulas employ SI units.

$$1\ \text{MeV}=1\times10^{6}\ \text{eV}=1\times10^{6}\times1.6\times10^{-19}\ \text{J}=1.6\times10^{-13}\ \text{J}$$

The classical relation between kinetic energy and speed is stated as

$$K=\frac12\,m\,v^{2}$$

where $$K$$ is the kinetic energy, $$m$$ the mass and $$v$$ the speed. Substituting $$K=1.6\times10^{-13}\ \text{J}$$ and $$m=1.6\times10^{-27}\ \text{kg}$$ (rest mass of the proton), we get

$$1.6\times10^{-13}=\frac12\,(1.6\times10^{-27})\,v^{2}$$

Multiplying both sides by 2 to remove the fraction,

$$2\times1.6\times10^{-13}=(1.6\times10^{-27})\,v^{2}$$

Notice that the factor $$1.6$$ appears on both sides, so it cancels:

$$2\times10^{-13}=10^{-27}\,v^{2}$$

Dividing by $$10^{-27}$$, we have

$$v^{2}=2\times10^{14}$$

Taking the square root,

$$v=\sqrt{2}\times10^{7}\ \text{m s}^{-1}\approx1.414\times10^{7}\ \text{m s}^{-1}$$

Now, the proton experiences an acceleration of $$10^{12}\ \text{m s}^{-2}$$ due to a magnetic field directed from west to east, while its velocity is south to north. The magnetic force formula is first stated:

$$\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}$$

The magnitude of this force when $$\mathbf{v}$$ and $$\mathbf{B}$$ are perpendicular is

$$F=q\,v\,B$$

According to Newton’s second law, this force produces an acceleration $$a$$, so

$$F=m\,a$$

Equating the two expressions for the force gives

$$m\,a=q\,v\,B$$

Solving for the magnetic field magnitude $$B$$,

$$B=\frac{m\,a}{q\,v}$$

We substitute $$m=1.6\times10^{-27}\ \text{kg}$$, $$a=10^{12}\ \text{m s}^{-2}$$, $$q=1.6\times10^{-19}\ \text{C}$$ and $$v=1.414\times10^{7}\ \text{m s}^{-1}$$:

$$B=\frac{(1.6\times10^{-27})(10^{12})}{(1.6\times10^{-19})(1.414\times10^{7})}$$

Simplifying the numerator,

$$1.6\times10^{-27}\times10^{12}=1.6\times10^{-15}$$

Simplifying the denominator,

$$(1.6\times10^{-19})(1.414\times10^{7})=1.6\times1.414\times10^{-12}=2.2624\times10^{-12}$$

Hence,

$$B=\frac{1.6\times10^{-15}}{2.2624\times10^{-12}}=\frac{1.6}{2.2624}\times10^{-3}\ \text{T}$$

Calculating the numerical ratio,

$$\frac{1.6}{2.2624}\approx0.707$$

Therefore,

$$B\approx0.707\times10^{-3}\ \text{T}=0.707\ \text{mT}$$

This value matches closely with $$0.71\ \text{mT}$$.

Hence, the correct answer is Option A.

We start with the definition of magnetic moment for a planar current loop. A charge $$q$$ completing one revolution in time period $$T$$ constitutes a current

$$I \;=\;\frac{q}{T}.$$

The particle moves in a circle of radius $$r$$ with speed $$v$$, so its period is obtained from the well-known relation $$T=\frac{\text{circumference}}{\text{speed}}$$ :

$$T \;=\;\frac{2\pi r}{v}.$$

Substituting this value of $$T$$ in the formula for current, we obtain

$$I \;=\;\frac{q}{\dfrac{2\pi r}{v}} \;=\;\frac{qv}{2\pi r}.$$

The area of the circular path is

$$A \;=\;\pi r^{2}.$$

By definition the magnitude of the magnetic moment is

$$\mu \;=\; I\,A.$$

Inserting the expressions for $$I$$ and $$A$$ gives

$$\mu \;=\;\left(\frac{qv}{2\pi r}\right)\!(\pi r^{2}) \;=\;\frac{qvr}{2}.$$

Its direction is perpendicular to the plane of the circle, along the unit vector $$\hat n$$ obtained from the right-hand rule for a positive charge. Hence in vector form

$$\vec\mu \;=\;\frac{qvr}{2}\,\hat n.$$

Next, we express $$r$$ in terms of the given magnetic field $$\vec B$$. Because the velocity is perpendicular to $$\vec B$$, the magnetic Lorentz force provides the necessary centripetal force. Writing this equality explicitly,

$$qvB \;=\;\frac{mv^{2}}{r}.$$

Solving for the radius,

$$r \;=\;\frac{mv}{qB}.$$

Substituting this value of $$r$$ back into $$\vec\mu$$, we obtain

$$\vec\mu \;=\;\frac{qv}{2}\left(\frac{mv}{qB}\right)\hat n \;=\;\frac{mv^{2}}{2B}\,\hat n.$$

The angular momentum $$\vec L$$ of the particle is $$\vec L = mvr\,\hat n$$, and for a positive charge $$\vec\mu$$ is parallel to $$\vec L$$, whereas for a negative charge it is antiparallel. The options given contain a negative sign, so the particle is to be regarded as negatively charged. Thus we write

$$\vec\mu \;=\;-\frac{mv^{2}}{2B}\,\hat B,$$

because $$\hat n$$ is opposite to $$\hat B$$ for a negative charge in this standard configuration. Since $$\hat B = \dfrac{\vec B}{B}$$, we finally have

$$\vec\mu \;=\;-\frac{mv^{2}}{2B^{2}}\;\vec B.$$

This matches Option D.

Hence, the correct answer is Option D.

We have a flat circular coil whose plane is kept vertical at the beginning, so the magnetic moment vector $$\vec \mu$$ is horizontal while the external uniform magnetic field $$\vec B$$ is vertical. Hence the initial angle between the two vectors is $$\theta_0 = 90^{\circ}$$.

The coil can rotate freely about one of its horizontal diameters. When the field is suddenly switched on, the magnetic torque tries to turn the coil so that $$\vec \mu$$ may align with $$\vec B$$. We are told that the coil actually rotates through an angle of $$60^{\circ}$$ about the horizontal diameter. Therefore the new angle between $$\vec \mu$$ and $$\vec B$$ becomes

$$\theta = 90^{\circ} - 60^{\circ} = 60^{\circ}.$$

The magnetic potential energy of a magnetic dipole in a uniform magnetic field is given first - we state the formula explicitly:

$$U = -\,\vec \mu \cdot \vec B = -\,\mu B \cos\theta.$$

Initially (at $$\theta_0 = 90^{\circ}$$),

$$U_0 = -\,\mu B \cos 90^{\circ} = -\,\mu B \,(0) = 0.$$

After the coil has turned (at $$\theta = 60^{\circ}$$),

$$U = -\,\mu B \cos 60^{\circ} = -\,\mu B \left(\dfrac12\right) = -\dfrac{\mu B}{2}.$$

Thus the change in potential energy experienced by the coil is

$$\Delta U = U - U_0 = \left(-\dfrac{\mu B}{2}\right) - 0 = -\dfrac{\mu B}{2}.$$

The negative sign shows that the energy of the system has decreased; that lost potential energy appears as rotational kinetic energy. By conservation of mechanical energy:

$$\dfrac12 I \omega^{2} = -\Delta U = \dfrac{\mu B}{2}.$$

We now insert the numerical values given in the problem:

Moment of inertia: $$I = 0.8 \text{ kg m}^2,$$
Magnetic moment: $$\mu = 20 \text{ A m}^2,$$
Magnetic field: $$B = 4 \text{ T}.$$

Substituting these values into the energy-conservation equation, we get

$$\dfrac12 \,(0.8)\,\omega^{2} = \dfrac{(20)(4)}{2}.$$

Simplifying the right-hand side first,

$$\dfrac{(20)(4)}{2} = \dfrac{80}{2} = 40,$$

so the equation becomes

$$0.4\,\omega^{2} = 40.$$

Solving step by step:

$$\omega^{2} = \dfrac{40}{0.4} = 100,$$

and therefore

$$\omega = \sqrt{100} = 10 \text{ rad s}^{-1}.$$

Hence, the correct answer is Option A.

We are asked to find the magnetic dipole moment of the iron rod after it is inserted in the solenoid and an electric current is sent through the turns. Let us introduce the symbols that will be used:

$$$V = 10^{-3}\,\text{m}^3 \qquad (\text{volume of the rod})$$$ $$$\mu_r = 1000 \qquad (\text{relative permeability of iron})$$$ $$$n = 10\,\text{turns cm}^{-1} \qquad (\text{number of turns per centimetre})$$$ $$$I = 0.5\,\text{A} \qquad (\text{current through the solenoid})$$$

First we convert the turn density into SI units (turns per metre). Since $$1\,\text{cm} = 10^{-2}\,\text{m},$$ we obtain

$$$n = 10\,\text{turns cm}^{-1} = 10 \times \frac{1}{10^{-2}}\,\text{turns m}^{-1} = 10 \times 100\,\text{turns m}^{-1} = 1000\,\text{turns m}^{-1}.$$$

Now we recall the definition of the magnetising field $$H$$ inside an ideal long solenoid. The standard formula is

$$H = n I,$$

where $$n$$ is the number of turns per metre and $$I$$ is the current in amperes. Substituting the known values we get

$$$H = (1000\,\text{turns m}^{-1})(0.5\,\text{A}) = 500\,\text{A m}^{-1}.$$$

Next we connect the magnetising field $$H$$ to the magnetisation $$M$$ of the iron core. For a linear magnetic material we use the relation

$$M = \chi_m H,$$

where $$\chi_m$$ is the magnetic susceptibility given by

$$\chi_m = \mu_r - 1.$$

For the present rod,

$$\chi_m = 1000 - 1 = 999.$$

Hence the magnetisation becomes

$$$M = 999 \times 500\,\text{A m}^{-1} = 499{,}500\,\text{A m}^{-1} = 4.995 \times 10^{5}\,\text{A m}^{-1}.$$$

The total magnetic dipole moment $$m$$ of the rod is obtained by multiplying the magnetisation $$M$$ by the volume $$V$$ (because magnetisation is magnetic moment per unit volume):

$$m = M V.$$

Substituting $$M = 4.995 \times 10^{5}\,\text{A m}^{-1}$$ and $$V = 10^{-3}\,\text{m}^3$$ gives

$$$m = (4.995 \times 10^{5}\,\text{A m}^{-1})(10^{-3}\,\text{m}^3) = 4.995 \times 10^{2}\,\text{A m}^2.$$$

In scientific notation this is essentially

$$m \approx 5 \times 10^{2}\,\text{A m}^2.$$

This numerical value exactly matches option B among the given choices.

Hence, the correct answer is Option B.

First, let us recall what the two keywords in the options mean.

Retentivity (also called remanence) is the residual value of magnetic induction $$B$$ left in the material when the magnetising field $$H$$ is reduced to zero. In simpler words, a material with large retentivity can hold on to its magnetism for a long time after the external field is removed.

Coercivity is the reverse magnetic field that must be applied to bring that residual induction back to zero. Thus a material with large coercivity needs a strong opposite field to demagnetise it; in other words, it strongly resists any attempt to lose its magnetisation.

Now we analyse the requirements for the two devices mentioned in the question.

For a permanent magnet, we want the magnet to remain magnetised under all ordinary circumstances. Therefore we require

$$\text{Retentivity : large}$$

so that a large residual magnetism is present, and

$$\text{Coercivity : large}$$

so that stray fields, heating, or mechanical shocks cannot easily demagnetise it.

For a transformer core, on the other hand, the magnetic field inside the core is made to reverse direction many times every second (50 Hz or 60 Hz). We need a material whose magnetisation follows the external field easily; otherwise energy will be wasted as heat. Hence, for a transformer we actually desire small coercivity and small retentivity, giving a narrow hysteresis loop and low hysteresis loss.

Scanning the given options, we see that Option D states

$$P : \text{large retentivity, large coercivity}$$

which is exactly the pair of properties we have just concluded to be essential for a permanent magnet. None of the other options simultaneously satisfies both conditions for the permanent magnet, and the statements given for the transformer in the other options are also not the ideal combination.

Hence, the correct answer is Option D.

For a straight, infinitely long wire carrying current $$I$$, the magnetic field at a perpendicular distance $$r$$ from the wire is given by the Biot-Savart result

$$\vec B=\frac{\mu_0 I}{2\pi r}\,\hat{\phi},$$

where $$\hat{\phi}$$ is the azimuthal (tangential) unit vector around the wire. In our situation the wire runs along the $$z$$-axis through the point $$(0,b,0)$$, so every field line lies in the $$xy$$-plane centred on that line.

Any current element $$I\,d\vec l$$ placed in a magnetic field experiences a differential force given by

$$d\vec F=I\;d\vec l\times\vec B.$$

The torque of this force about the $$z$$-axis is

$$d\tau_z=(\vec r\times d\vec F)_z=x\,dF_y-y\,dF_x,$$

with $$\vec r=(x,y,z)$$ measured from the origin.

The square loop of side $$2a$$ lies in the $$xz$$-plane ($$y=0$$) and is centred at the origin, so its four straight sides are

1. $$AB: (-a,0,-a)\to(a,0,-a)\ \ (\text{along }+x)\\ 2. BC: (a,0,-a)\to(a,0,a)\ \ (\text{along }+z)\\ 3. CD: (a,0,a)\to(-a,0,a)\ \ (\text{along }-x)\\ 4. DA: (-a,0,a)\to(-a,0,-a)\ (\text{along }-z).$$

Because every point of the loop has coordinate $$y=0$$, the distance of a generic point $$(x,0,z)$$ from the wire is

$$r=\sqrt{x^{2}+b^{2}},$$

and the magnetic field there is obtained by writing the azimuthal unit vector explicitly. The vector from the wire to the field point is $$\vec r=x\,\hat i-b\,\hat j,$$ so

$$\hat{\phi}=\frac{\hat k\times\vec r}{|\vec r|}=\frac{b\,\hat i+x\,\hat j}{r}.$$

Substituting in the Biot-Savart expression gives the field on the loop:

$$\vec B(x)=\frac{\mu_0 I}{2\pi r}\;\hat{\phi}= \frac{\mu_0 I}{2\pi}\;\frac{b\,\hat i+x\,\hat j}{x^{2}+b^{2}}.$$

Now we treat each side in turn.

Sides AB and CD (segments parallel to the $$x$$-axis)

On these sides $$d\vec l=\pm dx\,\hat i$$, so

$$d\vec F=I\,d\vec l\times\vec B =I\,dx\,\hat i\times\frac{\mu_0 I}{2\pi}\frac{b\,\hat i+x\,\hat j}{x^{2}+b^{2}} =\frac{\mu_0 I^{2}}{2\pi}\frac{x\,dx}{x^{2}+b^{2}}\;\hat k.$$ Because the force is purely along $$\hat k$$, both $$F_x$$ and $$F_y$$ are zero and therefore $$d\tau_z=0$$ on these two sides. They contribute no torque about the $$z$$-axis.

Side BC (segment parallel to the $$z$$-axis at $$x=+a$$)

Here $$d\vec l=dz\,\hat k$$ and $$x=a$$ is constant, so

$$d\vec F_{BC}=I\,dz\,\hat k\times \frac{\mu_0 I}{2\pi}\frac{b\,\hat i+a\,\hat j}{a^{2}+b^{2}} =\frac{\mu_0 I^{2}}{2\pi(a^{2}+b^{2})}\,dz\, \bigl[b\,(\hat k\times\hat i)+a\,(\hat k\times\hat j)\bigr] =\frac{\mu_0 I^{2}}{2\pi(a^{2}+b^{2})}\,dz\,(b\,\hat j-a\,\hat i).$$

Its components are $$F_x=-\dfrac{\mu_0 I^{2}a}{2\pi(a^{2}+b^{2})}\,dz$$ and $$F_y=\dfrac{\mu_0 I^{2}b}{2\pi(a^{2}+b^{2})}\,dz$$. The position vector on this side is $$\vec r=(a,0,z)$$, so

$$d\tau_z^{(BC)}=x\,dF_y-y\,dF_x=a\,dF_y =\frac{\mu_0 I^{2}a b}{2\pi(a^{2}+b^{2})}\,dz.$$

Integrating $$z$$ from $$-a$$ to $$+a$$ we obtain

$$\tau_z^{(BC)}=\int_{-a}^{a}d\tau_z^{(BC)} =\frac{\mu_0 I^{2}a b}{2\pi(a^{2}+b^{2})}\,(2a) =\frac{\mu_0 I^{2}a^{2} b}{\pi(a^{2}+b^{2})}.$$

Side DA (segment parallel to the $$z$$-axis at $$x=-a$$)

On this side the current runs in the negative $$z$$-direction, so we write $$d\vec l=-dz\,\hat k$$ while $$x=-a$$. The magnetic field there is

$$\vec B(-a)=\frac{\mu_0 I}{2\pi}\frac{b\,\hat i- a\,\hat j}{a^{2}+b^{2}}.$$

Hence

$$d\vec F_{DA}=I\,(-dz\,\hat k)\times \frac{\mu_0 I}{2\pi}\frac{b\,\hat i- a\,\hat j}{a^{2}+b^{2}} =-\frac{\mu_0 I^{2}}{2\pi(a^{2}+b^{2})}\,dz\, \bigl[b\,(\hat k\times\hat i)-a\,(\hat k\times\hat j)\bigr] =-\frac{\mu_0 I^{2}}{2\pi(a^{2}+b^{2})}\,dz\,(b\,\hat j+a\,\hat i).$$

The components are $$F_x=-\dfrac{\mu_0 I^{2}a}{2\pi(a^{2}+b^{2})}\,dz$$, $$F_y=-\dfrac{\mu_0 I^{2}b}{2\pi(a^{2}+b^{2})}\,dz$$. With $$\vec r=(-a,0,z)$$ we have

$$d\tau_z^{(DA)}=x\,dF_y-y\,dF_x=(-a)(dF_y) =\frac{\mu_0 I^{2}a b}{2\pi(a^{2}+b^{2})}\,dz,$$

exactly the same integrand as for side BC. Integrating from $$z=+a$$ back to $$z=-a$$ (the physical direction of current) still produces

$$\tau_z^{(DA)}=\frac{\mu_0 I^{2}a^{2} b}{\pi(a^{2}+b^{2})}.$$

Total torque about the $$z$$-axis

The two horizontal sides give zero torque, while each vertical side contributes the same positive amount found above, so

$$\tau_z=\tau_z^{(BC)}+\tau_z^{(DA)} =\frac{\mu_0 I^{2}a^{2} b}{\pi(a^{2}+b^{2})} +\frac{\mu_0 I^{2}a^{2} b}{\pi(a^{2}+b^{2})} =\frac{2\mu_0 I^{2}a^{2} b}{\pi(a^{2}+b^{2})}.$$

Therefore the magnitude of the torque acting on the square loop about the $$z$$-axis is

$$\boxed{\displaystyle \tau=\frac{2\mu_0 I^{2}a^{2} b}{\pi\!\left(a^{2}+b^{2}\right)}}.$$

Hence, the correct answer is Option B.

We have a square current-carrying loop of side $$2a$$ whose plane is the $$XZ$$-plane and whose centre is at the origin. Hence every point of the loop has coordinate $$y = 0$$. The loop carries current $$I$$. A very long straight conductor, also carrying the same current $$I$$, is parallel to the $$z$$-axis and passes through the fixed point $$(0,\, b,\, 0)$$ with $$b \gg a$$. Thus the straight wire is at a perpendicular distance $$b$$ from the centre of the square loop.

First we note the magnetic moment (also called magnetic dipole moment) of a current loop. For any closed loop the magnitude is given by the formula

$$m \;=\; I \, A,$$

where $$A$$ is the area enclosed by the loop and its direction is perpendicular to the plane of the loop according to the right-hand rule. Here the area of the square is

$$A \;=\; (2a)\times(2a) \;=\; 4a^{2}.$$

Hence the magnitude of the magnetic moment of the square loop is

$$m \;=\; I \,(4a^{2}) \;=\; 4 I a^{2}.$$

The loop lies in the $$XZ$$-plane, so its area vector (and therefore its magnetic-moment vector) points along the $$\pm y$$-axis. We shall simply denote its magnitude by $$m$$ and keep in mind that its direction is along $$\hat y$$ or $$-\hat y$$ depending on the sense of current; that sign will automatically be taken care of by the cross product.

Next we need the magnetic field produced by the long straight wire at a generic point of the loop. For an infinitely long wire the magnitude of the magnetic field at a perpendicular distance $$r$$ is given by the standard Biot-Savart result

$$B \;=\;\frac{\mu_{0} I}{2\pi r}.$$

The distance of the centre of the loop from the wire is exactly $$b$$, so the field at the centre due to the wire is

$$B_{0} \;=\;\frac{\mu_{0} I}{2\pi b}.$$

Because we are told that $$b \gg a$$, every point of the square loop is very nearly at the same distance $$b$$ from the wire; the fractional variation in distance is only of order $$a/b$$. Therefore, to an excellent approximation, we can treat the magnetic field over the whole loop as uniform and equal to $$B_{0}$$ in magnitude and direction. This simplification allows us to use the simple dipole-torque formula instead of integrating the force on each element.

The torque $$\vec\tau$$ acting on a current loop of magnetic moment $$\vec m$$ placed in a (uniform) magnetic field $$\vec B$$ is given by the vector relation

$$\vec\tau \;=\; \vec m \times \vec B.$$

The magnitude of the torque is therefore

$$\tau \;=\; m B \sin\theta,$$

where $$\theta$$ is the angle between $$\vec m$$ and $$\vec B$$. In our geometry the magnetic field of the straight wire is tangential to a circle centred on the wire; at the origin (centre of the square) that direction is along the $$+x$$-axis. Thus $$\vec B$$ is along $$\hat x$$ while $$\vec m$$ is along $$\pm\hat y$$, so the angle between them is exactly $$90^\circ$$ and $$\sin\theta = 1$$.

Putting everything together, the magnitude of the torque is

$$\tau \;=\; m B_{0} \;=\; \bigl(4 I a^{2}\bigr) \left(\frac{\mu_{0} I}{2\pi b}\right).$$

Multiplying the factors gives

$$\tau \;=\; \frac{4\,\mu_{0}\,I^{2}\,a^{2}}{2\pi b} \;=\; \frac{2\,\mu_{0}\,I^{2}\,a^{2}}{\pi b}.$$

This matches option C in the list provided.

Hence, the correct answer is Option 3.

We recall the torque formula for a current-carrying coil placed in a uniform magnetic field. The magnitude of the torque is given by

$$\tau = N I A B \sin\theta,$$

where $$N$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of each turn, $$B$$ is the magnetic-field strength, and $$\theta$$ is the angle between the magnetic-field direction and the normal to the plane of the coil.

Now the coil is stated to be parallel to the magnetic field. This makes the normal to the coil perpendicular to the field, giving $$\theta = 90^\circ$$ and hence $$\sin\theta = 1$$. Therefore the expression for torque simplifies to

$$\tau = N I A B.$$

We substitute the given numerical values. We have

$$N = 500, \qquad I = 0.5\ \text{A}, \qquad A = 3 \times 10^{-4}\ \text{m}^2, \qquad \tau = 1.5\ \text{N\,m}.$$

Putting these into the simplified torque formula gives

$$1.5 = 500 \times 0.5 \times \left(3 \times 10^{-4}\right) \times B.$$

First evaluate the product in the denominator step by step:

$$500 \times 0.5 = 250,$$

$$250 \times 3 = 750,$$

$$750 \times 10^{-4} = 7.5 \times 10^{-2} = 0.075.$$

So we have

$$1.5 = 0.075 \, B.$$

Solving for $$B$$, we divide both sides by 0.075:

$$B = \frac{1.5}{0.075} = 20.$$

Thus the required magnetic-field strength is

$$B = 20\ \text{T}.$$

So, the answer is $$20\ \text{T}$$.

We are asked to find the magnetic flux through the wire loop that follows the edges A → B → C → D → E → F → A, where the corner coordinates are

$$$A(0,0,0),\, B(5,0,0),\, C(5,5,0),\, D(0,5,0)$$$, $$E(0,5,5),\, F(0,0,5).$$

The magnetic field in the region is uniform and given by

$$\vec B = 3\hat i + 4\hat k\;\text{ T}.$$

The loop is not planar, so instead of splitting the surface into several rectangles, it is quicker to use a vector formula that gives the net area vector directly from the boundary curve. For any closed contour $$\mathcal C$$, an oriented surface bounded by that contour has area vector

$$\vec A \;=\;\frac12\oint_{\mathcal C}\vec r\times d\vec r,$$

where $$\vec r$$ is the position vector of a point on the contour and $$d\vec r$$ is the differential displacement along the contour. This follows from Stokes’ theorem and is valid no matter how the surface is bent; the result depends only on the boundary.

We therefore evaluate the line integral $$\displaystyle\oint\vec r\times d\vec r$$ one edge at a time, always moving in the given order A → B → C → D → E → F → A.

Edge AB: $$$\vec r=\langle x,0,0\rangle,\;0\le x\le5,\;d\vec r=\langle dx,0,0\rangle.$$$ So $$\vec r\times d\vec r=\vec0,$$ and the integral along AB is $$\vec0.$$

Edge BC: $$$\vec r=\langle5,y,0\rangle,\;0\le y\le5,\;d\vec r=\langle0,dy,0\rangle.$$$ Then $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\5&y&0\\0&dy&0\end{vmatrix}=5\,dy\,\hat k.$$$ Integrating, $$\int_{0}^{5}5\,dy\,\hat k=25\hat k.$$

Edge CD: $$$\vec r=\langle x,5,0\rangle,\;5\ge x\ge0,\;d\vec r=\langle dx,0,0\rangle.$$$ Here $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\x&5&0\\dx&0&0\end{vmatrix}=-5\,dx\,\hat k.$$$ Integrating from 5 down to 0, $$\int_{5}^{0}(-5)\,dx\,\hat k=25\hat k.$$

Edge DE: $$$\vec r=\langle0,5,z\rangle,\;0\le z\le5,\;d\vec r=\langle0,0,dz\rangle.$$$ Thus $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\0&5&z\\0&0&dz\end{vmatrix}=5\,dz\,\hat i.$$$ Integrating, $$\int_{0}^{5}5\,dz\,\hat i=25\hat i.$$

Edge EF: $$$\vec r=\langle0,y,5\rangle,\;5\ge y\ge0,\;d\vec r=\langle0,dy,0\rangle.$$$ Then $$$\vec r\times d\vec r=\begin{vmatrix}\hat i&\hat j&\hat k\\0&y&5\\0&dy&0\end{vmatrix}=-5\,dy\,\hat i.$$$ Integrating from 5 down to 0, $$\int_{5}^{0}(-5)\,dy\,\hat i=25\hat i.$$

Edge FA: $$$\vec r=\langle0,0,z\rangle,\;5\ge z\ge0,\;d\vec r=\langle0,0,dz\rangle.$$$ The cross‐product is zero, so the contribution is $$\vec0.$$

Adding all six contributions we get

$$$\oint\vec r\times d\vec r=(25+25)\hat i+(25+25)\hat k= \langle50,\,0,\,50\rangle.$$$

Dividing by 2, the area vector bounded by the loop is

$$$\vec A=\frac12\langle50,\,0,\,50\rangle=\langle25,\,0,\,25\rangle\text{ m}^2.$$$

Magnetic flux is defined as

$$\Phi=\vec B\cdot\vec A.$$

Substituting $$\vec B=\langle3,\,0,\,4\rangle\text{ T}$$ and $$\vec A=\langle25,\,0,\,25\rangle\text{ m}^2,$$ we have

$$$\Phi=(3)(25)+(0)(0)+(4)(25)=75+100=175\;\text{Wb}.$$$

So, the answer is $$175$$.

We first recall the relation between the restoring torque acting on a magnetic dipole and the angular acceleration it produces. A magnetic dipole of moment $$\mu$$ kept in a uniform magnetic field $$B$$ experiences a torque

$$\tau = \mu B\sin\theta,$$

where $$\theta$$ is the angle between $$\vec\mu$$ and $$\vec B$$. For very small oscillations, $$\sin\theta \approx \theta$$, so the torque becomes

$$\tau \;=\; -\,\mu B\,\theta.$$

(The minus sign only indicates that the torque is restoring.)

In rotational dynamics the torque is also related to the angular acceleration $$\alpha$$ by

$$\tau = I\alpha = I\frac{d^{2}\theta}{dt^{2}},$$

where $$I$$ is the moment of inertia about the axis of rotation. Combining the two expressions, we obtain

$$I\frac{d^{2}\theta}{dt^{2}} = -\,\mu B\,\theta.$$

This is the differential equation of simple harmonic motion, with the square of the angular frequency given by

$$\omega^{2} = \frac{\mu B}{I}.$$

Hence the time-period of small oscillations is

$$T = 2\pi\sqrt{\frac{I}{\mu B}}.$$

Now we analyse the two given bodies one by one.

Hoop (ring)
We are told that the mass of the hoop is $$M$$ and the radius is $$R$$. About its own axis the moment of inertia is well known:

$$I_h = MR^{2}.$$

The magnetic moment of the hoop is twice that of the solid cylinder, so if we denote the cylinder’s magnetic moment by $$\mu_c$$, then

$$\mu_h = 2\mu_c.$$

Substituting these values in the period formula, we find

$$T_h = 2\pi\sqrt{\frac{I_h}{\mu_h B}} = 2\pi\sqrt{\frac{MR^{2}}{(2\mu_c)B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Solid cylinder
For a solid cylinder of the same mass $$M$$ and radius $$R$$, the moment of inertia about its own axis is

$$I_c = \frac{1}{2}MR^{2}.$$

Its magnetic moment is simply $$\mu_c.$$
Putting these quantities into the period formula gives

$$T_c = 2\pi\sqrt{\frac{I_c}{\mu_c B}} = 2\pi\sqrt{\frac{\dfrac{1}{2}MR^{2}}{\mu_c B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

We see that the square-root expressions for $$T_h$$ and $$T_c$$ are identical:

$$T_h = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}} \quad \text{and} \quad T_c = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Therefore

$$T_h = T_c.$$

Hence, the correct answer is Option B.

We recall the general result for a current-carrying coil in a uniform magnetic field. A planar coil with $$N$$ turns, carrying current $$I$$, and having vector area $$\vec A$$ possesses a magnetic dipole moment $$\vec\mu$$ defined by the formula

$$\vec\mu = N\,I\,\vec A.$$

Here the vector $$\vec A$$ has magnitude equal to the geometrical area of the coil and its direction is perpendicular to the plane of the coil, obtained from the right-hand rule (curl the fingers in the sense of current, the thumb gives the direction of $$\vec A$$).

The torque $$\vec\tau$$ acting on a magnetic dipole in a uniform magnetic field $$\vec B$$ is given by the vector formula

$$\vec\tau = \vec\mu \times \vec B.$$

From this cross-product, the magnitude of the torque is

$$\tau = \mu\,B\,\sin\theta,$$

where $$\theta$$ is the angle between the magnetic moment $$\vec\mu$$ (or equivalently the area vector $$\vec A$$) and the magnetic field $$\vec B$$.

Now, for the circular coil described in the question:

• It has $$N$$ turns.
• Its radius is $$r$$, so its geometrical area (for one turn) is

$$A = \pi r^{2}.$$

Because the coil lies in the $$XZ$$-plane, its area vector (and hence $$\vec\mu$$) is directed along the $$\pm \hat{j}$$ (the $$Y$$) axis. The magnetic field is given as $$\vec B = B\,\hat{i},$$ i.e. along the $$X$$ axis. Therefore the angle $$\theta$$ between $$\vec\mu$$ (along $$\hat{j}$$) and $$\vec B$$ (along $$\hat{i}$$) is

$$\theta = 90^\circ,$$

for which $$\sin\theta = \sin 90^\circ = 1.$$

First, write the magnitude of the magnetic dipole moment:

$$\mu = N\,I\,A = N\,I\,(\pi r^{2}).$$

Substituting this value and $$\sin\theta = 1$$ into the torque formula, we have

$$\tau = \mu\,B\,\sin\theta = \bigl(N\,I\,\pi r^{2}\bigr)\,B\,(1).$$

Simplifying the factors, the magnitude of the torque becomes

$$\tau = B\,\pi r^{2}\,I\,N.$$

This matches exactly the expression given in Option A.

Hence, the correct answer is Option A.

For a paramagnetic substance we use Curie’s law, which states that the magnetic susceptibility $$\chi$$ varies inversely with the absolute temperature $$T$$:

$$\chi = \dfrac{C}{T},$$

where $$C$$ is Curie’s constant (a material‐specific constant).

Because the same specimen is observed at two different temperatures, the value of $$C$$ remains unchanged. Thus we may write for the two sets of observations

$$\chi_1 T_1 = C = \chi_2 T_2.$$

Rearranging to obtain the unknown susceptibility $$\chi_2$$ at temperature $$T_2$$ we get

$$\chi_2 = \chi_1 \dfrac{T_1}{T_2}.$$

The data provided are

Initial susceptibility: $$\chi_1 = 2.8 \times 10^{-4}.$$

Initial temperature: $$T_1 = 350 \text{ K}.$$

Required temperature: $$T_2 = 300 \text{ K}.$$

Substituting these values into the rearranged Curie’s law expression:

$$\chi_2 = \left(2.8 \times 10^{-4}\right)\dfrac{350}{300}.$$

First evaluate the fraction $$\dfrac{350}{300}$$:

$$\dfrac{350}{300} = \dfrac{35}{30} = \dfrac{7}{6} = 1.166\overline{6}.$$

Now multiply this factor with $$2.8 \times 10^{-4}$$:

$$\chi_2 = 2.8 \times 10^{-4} \times 1.166\overline{6}.$$

Carrying out the multiplication step by step:

$$2.8 \times 1.166\overline{6} = 3.266\overline{6}.$$

Hence

$$\chi_2 \approx 3.266\overline{6} \times 10^{-4}.$$

Rounding to three significant figures,

$$\chi_2 \approx 3.267 \times 10^{-4}.$$

This value matches the entry in Option C.

Hence, the correct answer is Option C.

We begin by recalling that the magnetic moment of a uniformly magnetised body is related to its magnetisation by

$$\vec M \;=\; \frac{\text{total magnetic moment}}{\text{volume of the body}}\;.$$

Here the substance is a cube whose edge length is given as $$1\ \text{cm}$$. Converting this length into SI units, we have

$$1\ \text{cm}=1\times10^{-2}\ \text{m}=0.01\ \text{m}\,.$$

Now the volume of a cube is the cube of its side, so

$$V=(0.01\ \text{m})^{3}=0.01^{3}\ \text{m}^{3}=1\times10^{-6}\ \text{m}^{3}\,.$$

The total magnetic dipole moment of the cube is given as $$20\times10^{-6}\ \text{J/T}$$. Remembering that $$1\ \text{J/T}=1\ \text{A·m}^{2}$$, we can write

$$\mu =20\times10^{-6}\ \text{A·m}^{2}\,.$$

Substituting the values of $$\mu$$ and $$V$$ in the definition of magnetisation, we get

$$M=\frac{\mu}{V}=\frac{20\times10^{-6}\ \text{A·m}^{2}}{1\times10^{-6}\ \text{m}^{3}}=20\ \text{A/m}\,.$$

Next, magnetic susceptibility $$\chi_{m}$$ is defined as the ratio of magnetisation $$M$$ to the applied magnetic intensity $$H$$, that is,

$$\chi_{m}=\frac{M}{H}\,.$$

The applied magnetic intensity is given as $$60\times10^{3}\ \text{A/m}=60000\ \text{A/m}$$. Hence, substituting the known values,

$$\chi_{m}=\frac{20\ \text{A/m}}{60000\ \text{A/m}} =\frac{20}{60000} =\frac{2}{6000} =\frac{1}{3000} =3.33\times10^{-4}\,.$$

So we obtain

$$\chi_{m}\;\approx\;3.3\times10^{-4}\,.$$

Hence, the correct answer is Option D.

We are told that the particle carries the same charge as an electron, so its charge is

$$q = 1.6 \times 10^{-19}\ \text{C}.$$

Under only the magnetic field, the particle executes uniform circular motion of radius

$$r = 0.5\ \text{cm} = 0.5 \times 10^{-2}\ \text{m} = 5 \times 10^{-3}\ \text{m}.$$

The magnetic field magnitude is

$$B = 0.5\ \text{T}.$$

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force supplies the centripetal force. We first state this relationship:

Magnetic force: $$F_B = q v B.$$ Centripetal force: $$F_C = \dfrac{m v^{2}}{r}.$$

Because the motion is circular under the magnetic field alone, these two forces are equal:

$$q v B = \dfrac{m v^{2}}{r}.$$

We can cancel one factor of the speed $$v$$ from both sides to find an expression for $$v$$:

$$q B = \dfrac{m v}{r}\quad \Longrightarrow\quad v = \dfrac{q B r}{m}.$$

Next, an electric field of magnitude

$$E = 100\ \text{V m}^{-1}$$

is switched on simultaneously with the magnetic field. The problem states that, under the simultaneous action of $$ \mathbf{E} $$ and $$ \mathbf{B} $$, the particle now moves in a straight line. A straight-line trajectory in crossed, perpendicular electric and magnetic fields implies that the net force is zero. The electric force and magnetic force must therefore cancel:

Electric force: $$F_E = q E.$$ Magnetic force (already written): $$F_B = q v B.$$

For cancellation, we must have

$$q E = q v B.$$

Because $$q \neq 0$$, we divide both sides by $$q$$ to obtain the standard crossed-fields condition:

$$E = v B \quad\Longrightarrow\quad v = \dfrac{E}{B}.$$

We now have two independent expressions for the speed $$v$$:

$$v = \dfrac{q B r}{m}\qquad\text{and}\qquad v = \dfrac{E}{B}.$$

Since both describe the same speed, we equate them:

$$\dfrac{q B r}{m} = \dfrac{E}{B}.$$

Solving this equation for the mass $$m$$, we first cross-multiply:

$$q B r \, B = m E.$$

This simplifies to

$$m = \dfrac{q B^{2} r}{E}.$$

Now we substitute the numerical values, keeping all factors explicit:

$$m = \dfrac{(1.6 \times 10^{-19}\ \text{C})\,(0.5\ \text{T})^{2}\,(5 \times 10^{-3}\ \text{m})}{100\ \text{V m}^{-1}}.$$

First calculate the square of the magnetic field:

$$(0.5\ \text{T})^{2} = 0.25\ \text{T}^{2}.$$

Multiply $$q$$ and $$B^{2}$$:

$$1.6 \times 10^{-19} \times 0.25 = 0.4 \times 10^{-19} = 4.0 \times 10^{-20}.$$

Now include the radius $$r$$:

$$4.0 \times 10^{-20} \times 5 \times 10^{-3} = 20.0 \times 10^{-23} = 2.0 \times 10^{-22}.$$

Finally, divide by the electric field magnitude $$E = 100 = 1.0 \times 10^{2}$$:

$$m = \dfrac{2.0 \times 10^{-22}}{1.0 \times 10^{2}} = 2.0 \times 10^{-24}\ \text{kg}.$$

This value exactly matches option C.

Hence, the correct answer is Option C.

We begin by recalling the motion of a charged particle in a uniform magnetic field. When a particle of mass $$m$$ and charge $$q$$ enters a field $$B$$ with velocity $$v$$ perpendicular to the field, it moves in a circle of radius

$$r=\dfrac{mv}{qB}\,.$$

Next, because each particle starts from rest and is accelerated through the same potential difference $$V$$, the work-energy theorem gives the kinetic energy gained:

$$qV=\dfrac12\,mv^{2}\;.$$

Solving this energy equation for the speed, we obtain

$$v=\sqrt{\dfrac{2qV}{m}}\;.$$

Substituting this value of $$v$$ into the formula for the radius, we arrive at

$$r=\dfrac{m}{qB}\,\sqrt{\dfrac{2qV}{m}} =\dfrac{1}{B}\,\sqrt{\dfrac{2mV}{q}}\;.$$

Thus, for any particle accelerated through the same potential difference $$V$$ and entering the same magnetic field $$B$$, the radius is proportional to $$\sqrt{\dfrac{m}{q}}$$.

Let us now evaluate this expression for the two given particles.

For the proton: its mass is $$m_{p}=m$$ and its charge is $$q_{p}=e$$. Therefore

$$r_{p}=\dfrac1B\,\sqrt{\dfrac{2mV}{e}}\;.$$

For the $$\alpha$$-particle: its mass is $$m_{\alpha}=4m$$ and its charge is $$q_{\alpha}=2e$$. Hence

$$r_{\alpha}=\dfrac1B\,\sqrt{\dfrac{2(4m)V}{2e}} =\dfrac1B\,\sqrt{\dfrac{8mV}{2e}} =\dfrac1B\,\sqrt{\dfrac{4mV}{e}} =\dfrac1B\;\;2\,\sqrt{\dfrac{mV}{e}}\;.$$

We now write the ratio of the two radii:

$$\dfrac{r_{p}}{r_{\alpha}} =\dfrac{\displaystyle \dfrac1B\,\sqrt{\dfrac{2mV}{e}}} {\displaystyle \dfrac1B\,2\,\sqrt{\dfrac{mV}{e}}} =\dfrac{\sqrt{2}}{2} =\dfrac1{\sqrt{2}}\;.$$

This gives

$$r_{p}:r_{\alpha}=1:\sqrt{2}\;.$$

Hence, the correct answer is Option B.

Given: $$K = 100\text{ eV} = 100 \times 1.6 \times 10^{-19}\text{ J}$$, $$B = 1.5 \times 10^{-3}\text{ T}$$, $$x = 2\text{ cm} = 0.02\text{ m}$$, $$L = 8\text{ cm} = 0.08\text{ m}$$

$$R = \frac{\sqrt{2mK}}{qB} = \frac{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}}{1.6 \times 10^{-19} \times 1.5 \times 10^{-3}} = \frac{5.4 \times 10^{-24}}{2.4 \times 10^{-22}} = 0.0225\text{ m} = 2.25\text{ cm}$$

$$\sin\theta = \frac{x}{R} = \frac{2}{2.25} = \frac{8}{9}$$

$$\tan\theta = \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} = \frac{8/9}{\sqrt{17}/9} = \frac{8}{\sqrt{17}}$$

Total horizontal distance from the exit point of the magnetic field to the screen: $$x' = L - x = 8 - 2 = 6\text{ cm}$$

Total vertical deflection ($$d$$) split into curved region ($$d_1$$) and straight region ($$d_2$$):

$$d_1 = R(1 - \cos\theta) = R\left(1 - \frac{\sqrt{17}}{9}\right) = 2.25 \times \left(1 - \frac{4.123}{9}\right) = 2.25 \times 0.5419 = 1.22\text{ cm}$$

$$d_2 = x'\tan\theta = 6 \times \frac{8}{\sqrt{17}} = \frac{48}{4.123} = 11.64\text{ cm}$$

$$d = d_1 + d_2 = 1.22 + 11.64 = 12.86\text{ cm} \approx 12.87\text{ cm}$$

$$p = \sqrt{2m eV}$$

$$r = \frac{p}{eB} = \frac{\sqrt{2m eV}}{eB} = \frac{1}{B}\sqrt{\frac{2mV}{e}}$$

$$r = \frac{1}{0.1}\sqrt{\frac{2 \times 9.1 \times 10^{-31} \times 500}{1.6 \times 10^{-19}}}$$

$$r = 10 \times \sqrt{\frac{9100 \times 10^{-31}}{1.6 \times 10^{-19}}} = 10 \times \sqrt{5687.5 \times 10^{-12}}$$

$$r \approx 10 \times 75.41 \times 10^{-6} = 7.54 \times 10^{-4}\ \text{m}$$

Step-by-Step Solution

1. Condition for Zero Net Force on Wire C

For a third parallel current-carrying wire $$C$$ to experience a net magnetic force of zero, the magnetic fields produced by wire $$A$$ ($$B_1$$) and wire $$B$$ ($$B_2$$) must be equal in magnitude but opposite in direction at the position of wire $$C$$.

$$F_{\text{net}} = 0 \implies B_1 = B_2$$

The magnetic field at a perpendicular distance $$r$$ from a long straight wire carrying current $$I$$ is given by:

$$B = \frac{\mu_0 I}{2\pi r}$$

2. Analysis of Directonal Fields (Regions)

Let's analyze the directions of the currents:

• Wire $$A$$ carries current $$I_1$$ upwards.
• Wire $$B$$ carries current $$I_2$$ downwards (opposite directions).
• Between the wires ($$0 < x < d$$): According to the right-hand grip rule, both wires produce magnetic fields pointing into (or out of) the page in the space between them. Since the fields point in the same direction, they reinforce each other, and the net field can never be zero in this region.
• Outside the wires ($$x < 0$$ or $$x > d$$): The magnetic fields point in opposite directions. Therefore, the net force can only be zero outside the two wires.
• $$x = \frac{I_1 d}{I_2 - I_1}$$ to the left of wire A (if $$I_2 > I_1$$)
• $$x = \frac{I_1 d}{I_1 - I_2}$$ to the right of wire B (if $$I_1 > I_2$$)

Because the currents are in opposite directions:

The exact valid region depends entirely on which current has a smaller magnitude, as wire $$C$$ must be placed closer to the weaker current to balance the stronger current's field.

3. Setting Up the Balance Equations

Case 1: Wire C is placed to the left of Wire A ($$x$$ is measured from A to the left)

Let $$x$$ be the distance from wire $$A$$ to the left. The distance from wire $$B$$ becomes $$(d + x)$$.

Equating the field magnitudes:

$$\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (d + x)}$$

Cancel out the common constant factors $$\frac{\mu_0}{2\pi}$$:

$$\frac{I_1}{x} = \frac{I_2}{d + x}$$

Cross-multiply and solve for $$x$$:

$$I_1(d + x) = I_2 x$$

$$I_1 d + I_1 x = I_2 x$$

$$I_1 d = (I_2 - I_1)x$$

$$x = \frac{I_1 d}{I_2 - I_1}$$

(This solution is physically valid only if $$I_2 > I_1$$.)

Case 2: Wire C is placed to the right of Wire B ($$x$$ is measured from A to the right)

If wire $$C$$ is placed to the right of wire $$B$$ at a distance $$x$$ from wire $$A$$, its distance from wire $$B$$ is $$(x - d)$$.

Equating the fields:

$$\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (x - d)} \implies \frac{I_1}{x} = \frac{I_2}{x - d}$$

Cross-multiply and solve for $$x$$:

$$I_1(x - d) = I_2 x$$

$$I_1 x - I_1 d = I_2 x$$

$$(I_1 - I_2)x = I_1 d$$

$$x = \frac{I_1 d}{I_1 - I_2}$$

(This solution is physically valid only if $$I_1 > I_2$$.)

Final Answer

The possible values of $$x$$ depend on the relative magnitudes of the currents:

We have a magnetic dipole whose magnetic moment vector is given as $$\vec m = 10^{-2}\,\hat i\;{\rm A\,m^2}.$$

The external magnetic field is time-dependent and is specified by

$$\vec B(t)=B\cos\omega t\;\hat i,$$

where the numerical values are $$B = 1\ {\rm T}$$ and $$\omega = 0.125\ {\rm rad\,s^{-1}}.$$

At the instant $$t = 1\ {\rm s}$$ the field becomes

$$\vec B(1)=1\cos(0.125\times1)\;\hat i =\cos(0.125)\;\hat i\ {\rm T}.$$

The angle $$0.125$$ rad is small, so

$$\cos(0.125)\approx 0.9922.$$

Hence the field magnitude at this instant is approximately

$$B_1 = 0.9922\ {\rm T}.$$

The formula for the potential energy of a magnetic dipole in a magnetic field is

$$U = -\vec m\cdot\vec B.$$

Initially the dipole is parallel to the field, so the angle between $$\vec m$$ and $$\vec B$$ is $$0^\circ$$. Therefore

$$U_i = -mB_1\cos 0^\circ = -mB_1.$$

To reverse the dipole, we make it point opposite to the field; the angle becomes $$180^\circ$$. Then

$$U_f = -mB_1\cos 180^\circ = -mB_1(-1) = +mB_1.$$

The work done by an external agent equals the increase in potential energy:

$$W = U_f - U_i = \left(+mB_1\right) - \left(-mB_1\right) = 2mB_1.$$

Substituting $$m = 10^{-2}\ {\rm A\,m^2}$$ and $$B_1 = 0.9922\ {\rm T}$$, we get

$$W = 2 \times 10^{-2} \times 0.9922 = 0.019844\ {\rm J} \approx 0.02\ {\rm J}.$$

Hence, the correct answer is Option B.

We begin by recalling the theory of the small oscillations of a magnetic (compass) needle. When the needle is displaced slightly in the horizontal plane, the restoring torque is $$\tau = M B_H \sin\theta \approx M B_H \theta$$ for small $$\theta,$$ where $$M$$ is the magnetic moment and $$B_H$$ is the horizontal component of the earth’s magnetic field. For a body executing small-angle simple harmonic motion, the angular frequency is given by the standard formula

$$\omega = \sqrt{\dfrac{\text{restoring torque per unit angle}}{\text{moment of inertia}}} = \sqrt{\dfrac{M B_H}{I}}.$$

The ordinary (linear) frequency is $$f = \dfrac{\omega}{2\pi},$$ so we have

$$f = \dfrac{1}{2\pi}\sqrt{\dfrac{M}{I}}\,\sqrt{B_H}.$$

Everything except $$B_H$$ is constant for the same needle; hence

$$f \propto \sqrt{B_H} \quad\Longrightarrow\quad B_H \propto f^2.$$

Now we convert the given “oscillations per minute” to “oscillations per second” (hertz):

At the first place, the needle makes 30 oscillations in 60 s, so

$$f_1 = \dfrac{30}{60} = 0.50\ \text{Hz}.$$

At the second place, the needle makes 40 oscillations in 60 s, so

$$f_2 = \dfrac{40}{60} = 0.667\ \text{Hz}.$$

Using $$B_H \propto f^2,$$ we obtain

$$\dfrac{B_{H1}}{B_{H2}} = \left(\dfrac{f_1}{f_2}\right)^2 = \left(\dfrac{0.50}{0.667}\right)^2 = \left(0.75\right)^2 = 0.5625.$$

However, the question asks for the ratio of the total earth’s magnetic fields $$B_1$$ and $$B_2,$$ not just their horizontal components. The total field is related to its horizontal component by the well-known relation

$$B_H = B\cos\delta,$$

where $$\delta$$ is the dip (angle of inclination). Thus

$$B = \dfrac{B_H}{\cos\delta}.$$

So, for the two places,

$$\dfrac{B_1}{B_2} = \dfrac{B_{H1}/\cos\delta_1}{B_{H2}/\cos\delta_2} = \dfrac{B_{H1}}{B_{H2}}\;\dfrac{\cos\delta_2}{\cos\delta_1}.$$

The dips are $$\delta_1 = 45^\circ$$ and $$\delta_2 = 30^\circ.$$ We substitute their cosines:

$$\cos45^\circ = \dfrac{1}{\sqrt2} \approx 0.7071,\qquad \cos30^\circ = \dfrac{\sqrt3}{2} \approx 0.8660.$$

Therefore,

$$\dfrac{\cos\delta_2}{\cos\delta_1} = \dfrac{0.8660}{0.7071} \approx 1.2247.$$

Multiplying this with the earlier ratio of horizontal components, we get

$$\dfrac{B_1}{B_2} = 0.5625 \times 1.2247 \approx 0.688.$$

This value is very close to 0.7. Among the given options, the one that matches is 0.7.

Hence, the correct answer is Option D.

We consider a uniform magnetic field that is perpendicular to the plane of motion of the charged particle. For a charged particle of charge $$q$$ and mass $$m$$ moving with speed $$v$$ in a circle of radius $$r$$, the magnetic Lorentz force provides the required centripetal force. The relation is

$$q\,v\,B = \dfrac{m v^{2}}{r}$$

First we isolate the radius $$r$$. Cancelling one factor of $$v$$ from both sides, we obtain

$$q B = \dfrac{m v}{r} \quad \Longrightarrow \quad r = \dfrac{m v}{q B}$$

The question tells us that the proton, the electron, and the Helium nucleus all possess the same kinetic energy, which we denote by $$E$$. The kinetic energy of a particle is given by

$$E = \dfrac{1}{2} m v^{2}$$

Solving this expression for the speed $$v$$, we have

$$v = \sqrt{\dfrac{2E}{m}}$$

We now substitute this expression for $$v$$ into the formula for the radius:

$$r = \dfrac{m}{q B} \,\sqrt{\dfrac{2E}{m}}

= \dfrac{\sqrt{2 E}\, \sqrt{m}}{q B}$$

The factors $$\sqrt{2E}$$ and $$B$$ are common to all three particles because the energy and the magnetic field are the same for each. Hence, for comparison of radii, we may write the proportionality

$$r \;\propto\; \dfrac{\sqrt{m}}{q}$$

Therefore, the ordering of the radii depends solely on the ratio $$\sqrt{m}/q$$ for the proton, the electron, and the Helium nucleus. We list their masses and charges (magnitude of charge only, because radius is a positive quantity):

• Electron: $$m = m_e,\; q = e$$
• Proton: $$m = m_p,\; q = e$$
• Helium nucleus (alpha particle): $$m = 4 m_p,\; q = 2e$$

Now we compute the proportional factors:

$$

\begin{aligned}

r_e &\propto \dfrac{\sqrt{m_e}}{e},\\[4pt]

r_p &\propto \dfrac{\sqrt{m_p}}{e},\\[4pt]

r_{He} &\propto \dfrac{\sqrt{4 m_p}}{2e}

= \dfrac{2\sqrt{m_p}}{2e}

= \dfrac{\sqrt{m_p}}{e}.

\end{aligned}

$$

We immediately observe that

$$r_{He} \propto \dfrac{\sqrt{m_p}}{e} = r_p.$$

Thus $$r_{He} = r_p.$$ To compare with the electron, note that the electron’s mass is enormously smaller than the proton’s mass, i.e.

$$m_e \ll m_p \quad\Longrightarrow\quad \sqrt{m_e} \ll \sqrt{m_p}.$$

Consequently

$$r_e \propto \dfrac{\sqrt{m_e}}{e} \ll \dfrac{\sqrt{m_p}}{e} \propto r_p = r_{He}.$$

Collecting these results, we obtain the order

$$r_e \lt r_p = r_{He}.$$

This ordering corresponds to Option B.

Hence, the correct answer is Option B.

Axial segments:  $$B_{LP} = B_{MQ} = 0$$

Perpendicular segments at origin:  $$\vec{B}_{PS} = \frac{\mu_0 I}{4\pi d} (-\hat{k})$$

$$\vec{B}_{QN} = \frac{\mu_0 I}{4\pi d} (-\hat{k})$$

Total magnetic field:

$$\vec{B}_{\text{net}} = \vec{B}_{PS} + \vec{B}_{QN} = \frac{\mu_0 I}{2\pi d} (-\hat{k})$$

$$10^{-4} = \frac{(4\pi \times 10^{-7}) \cdot I}{2\pi \times 0.04} = 5 \times 10^{-6} I \implies I = 20\text{ A}$$

Perpendicular distance ($$d$$) from wire to point P: $$d = \sqrt{5^2 - 3^2} = 4\text{ cm} = 0.04\text{ m}$$

Angles from perpendicular to wire ends: $$\sin\theta_1 = \sin\theta_2 = \frac{3}{5}$$

Magnetic field calculation: $$B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$$

$$B = \frac{10^{-7} \times 5}{0.04} \left(\frac{3}{5} + \frac{3}{5}\right) = \frac{5 \times 10^{-7}}{0.04} \times \frac{6}{5} = \frac{6 \times 10^{-7}}{0.04} = 1.5 \times 10^{-5}\text{ T}$$

Let the current flowing through both wires be $$I$$ and let the total length of each wire be $$L$$. Both wires are identical, so they have the same length.

First we take the wire that is bent into a single circular loop. If the radius of this loop is $$R_1$$, then its circumference must equal the length of the wire. Therefore, using the basic relation “circumference = $$2\pi\times\text{radius}$$”, we have

$$2\pi R_1 = L \;\;\Longrightarrow\;\; R_1 = \frac{L}{2\pi}.$$

The magnetic field at the centre of a single circular loop carrying current $$I$$ is given by the standard formula

$$B = \frac{\mu_0 I}{2R}.$$

Substituting $$R = R_1$$, we obtain the field at the centre of this loop:

$$B_L = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2\left(\dfrac{L}{2\pi}\right)} = \frac{\mu_0 I}{2}\cdot\frac{2\pi}{L} = \frac{\mu_0 I\pi}{L}.$$

Now consider the second wire that is wound into a circular coil of $$N$$ identical turns. Let the radius of each turn be $$R_2$$. Because there are $$N$$ turns, the total length consumed is $$N$$ times the circumference of one turn. Hence

$$N\,(2\pi R_2) = L \;\;\Longrightarrow\;\; R_2 = \frac{L}{2\pi N}.$$

The magnetic field at the centre of a circular coil having $$N$$ turns is $$N$$ times the field produced by one turn (all turns are coaxial and their fields add). Therefore, using the same single-turn formula but multiplied by $$N$$, we have

$$B_C = N\left(\frac{\mu_0 I}{2R_2}\right) = \frac{\mu_0 N I}{2R_2}.$$

Substituting $$R_2 = \dfrac{L}{2\pi N}$$ into the above expression gives

$$B_C = \frac{\mu_0 N I}{2\left(\dfrac{L}{2\pi N}\right)} = \frac{\mu_0 N I}{2}\cdot\frac{2\pi N}{L} = \frac{\mu_0 I\pi N^2}{L}.$$

We now form the required ratio $$\dfrac{B_L}{B_C}$$:

$$\frac{B_L}{B_C} = \frac{\dfrac{\mu_0 I\pi}{L}} {\dfrac{\mu_0 I\pi N^2}{L}} = \frac{\mu_0 I\pi}{L}\cdot\frac{L}{\mu_0 I\pi N^2} = \frac{1}{N^2}.$$

Hence, the correct answer is Option A.

We have a bar magnet placed completely inside a current-carrying solenoid. In such a situation the magnetic field acting on the magnet is exactly the field produced by the solenoid itself. To make the magnet lose all its magnetisation, this external field must be equal in magnitude (but opposite in sense) to the magnet’s coercive field. Therefore the coercivity $$H_c$$ of the bar magnet is numerically the same as the magnetising field intensity $$H$$ created by the solenoid.

First we recall the standard formula for the field inside a long solenoid. For a solenoid having $$N$$ turns, length $$\ell$$, and carrying a current $$I$$,

$$B = \mu_0 \, n \, I,$$

where $$n$$ is the number of turns per unit length, $$n = \dfrac{N}{\ell}$$, and $$\mu_0$$ is the permeability of free space. The magnetic field intensity $$H$$ is related to the flux-density $$B$$ by

$$H = \dfrac{B}{\mu_0}.$$

Substituting $$B = \mu_0 n I$$ into this relation gives

$$H = \dfrac{\mu_0 n I}{\mu_0} = n I.$$

Thus, inside a long solenoid,

$$H = n I.$$

Now we substitute the numerical data given in the question. The solenoid has

$$N = 100 \text{ turns}, \qquad \ell = 0.2 \text{ m}.$$

So the turn density is

$$n = \dfrac{N}{\ell} = \dfrac{100}{0.2} = 500 \text{ turns per metre}.$$

The current flowing is

$$I = 5.2 \text{ A}.$$

Therefore the field intensity generated is

$$H = n I = 500 \times 5.2 = 2600 \text{ A/m}.$$

Since this field exactly cancels the magnet’s residual magnetism, the coercivity of the bar magnet is

$$H_c = 2600 \text{ A/m}.$$

Hence, the correct answer is Option 2.

The coil of the moving-coil galvanometer reaches equilibrium when the magnetic torque trying to turn the coil is exactly balanced by the restoring (torsional) torque of the suspension.

First, we state the two torques.

Magnetic torque on a current-carrying coil is given by the formula $$\tau_{\text{mag}} = n\,I\,A\,B\,\sin\phi,$$ where $$n$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of each turn, $$B$$ is the magnetic-field magnitude, and $$\phi$$ is the angle between the normal to the coil and the magnetic field.

The problem tells us that the magnetic field is parallel to the plane of the coil. Hence the normal to the coil is perpendicular to the field, so $$\phi = 90^\circ$$ and $$\sin\phi = \sin 90^\circ = 1.$$ Therefore, the magnetic torque reduces to $$\tau_{\text{mag}} = n\,I\,A\,B.$$

The restoring torque supplied by the torsion band obeys Hooke’s law for angular twist: $$\tau_{\text{rest}} = C\,\theta,$$ where $$C$$ is the torsion (restoring) constant and $$\theta$$ is the angular deflection in radians.

At equilibrium we have $$\tau_{\text{mag}} = \tau_{\text{rest}}.$$ So, $$n\,I\,A\,B = C\,\theta.$$

We need to isolate $$B$$:

$$B = \frac{C\,\theta}{n\,I\,A}.$$

Now we substitute every given quantity, converting each into SI units:

Number of turns: $$n = 175.$$

Current: $$I = 1\ \text{mA} = 1 \times 10^{-3}\ \text{A}.$$

Area of the coil: $$A = 1\ \text{cm}^2 = 1 \times 10^{-4}\ \text{m}^2.$$ (1 cm = 10⁻² m, so 1 cm² = 10⁻⁴ m².)

Torsion constant: $$C = 10^{-6}\ \text{N m rad}^{-1}.$$

Angular deflection: $$\theta = 1^\circ = \left(\frac{\pi}{180}\right)\ \text{rad} \approx 0.01745\ \text{rad}.$$

Substituting these numbers into the expression for $$B$$ gives

$$B = \frac{C\,\theta}{n\,I\,A} = \frac{(10^{-6})\,(0.01745)}{(175)\,(1 \times 10^{-3})\,(1 \times 10^{-4})}.$$

We now calculate the numerator:

$$C\,\theta = 10^{-6} \times 0.01745 = 1.745 \times 10^{-8}.$$

Next, the denominator:

$$n\,I\,A = 175 \times (1 \times 10^{-3}) \times (1 \times 10^{-4}) = 175 \times 10^{-7} = 1.75 \times 10^{-5}.$$

Dividing numerator by denominator: