JEE Magnetism & Magnetic Materials Questions

$$\text{Given: } \frac{\Delta l}{l} = 1\%,\ \frac{\Delta d}{d} = 1\%$$

$$B \propto \frac{l}{d^3}$$

$$\frac{\Delta B}{B} = \frac{\Delta l}{l} + 3\left(\frac{\Delta d}{d}\right)$$

$$\frac{\Delta B}{B}\% = 1\% + 3(1\%) = 4\%$$

We need to find the shunt resistance for a galvanometer.

G = 30 Ω, $$I_g$$ = 20 mA = 0.02 A, I = 3 A

$$S = \frac{GI_g}{I - I_g} = \frac{30 \times 0.02}{3 - 0.02} = \frac{0.6}{2.98} = \frac{60}{298} = \frac{30}{149}$$

So $$S = \frac{30}{149}$$ Ω, meaning X = 149.

The correct answer is Option 2: 149.

A magnetic dipole with dipole moment $$M$$ in a uniform magnetic field $$B$$ experiences a torque $$\tau = MB\sin\theta$$ and has potential energy $$U = -MB\cos\theta$$.

Given $$\tau = 80\sqrt{3}$$ N m and $$\theta = 60°$$:

$$MB\sin 60° = 80\sqrt{3}$$

$$MB \cdot \dfrac{\sqrt{3}}{2} = 80\sqrt{3}$$

$$MB = 160$$

The potential energy is: $$U = -MB\cos 60° = -160 \cdot \dfrac{1}{2} = -80$$ J

Hence, the correct answer is Option D.

Voltage sensitivity $$S_V$$ of a moving-coil galvanometer is defined as the deflection produced per unit applied voltage.
If the torsional constant of the suspension spring is $$k$$, then

Current sensitivity $$S_I$$ is $$\displaystyle S_I = \frac{NAB}{k} \quad -(1)$$
where
$$N$$ = number of turns, $$A$$ = area of each turn, $$B$$ = magnetic field.

Voltage sensitivity is obtained by dividing current sensitivity by the total resistance $$R$$ of the coil:

$$\displaystyle S_V = \frac{S_I}{R} = \frac{NAB}{kR} \quad -(2)$$

Because both coils have the same spring, the torsional constant $$k$$ is identical for them and cancels in the ratio.

For coil $$M_1$$:
$$N_1 = 15$$, $$A_1 = 3.6 \times 10^{-3}\,\text{m}^2$$, $$B_1 = 0.25\,\text{T}$$, $$R_1 = 5\,\Omega$$

For coil $$M_2$$:
$$N_2 = 21$$, $$A_2 = 1.8 \times 10^{-3}\,\text{m}^2$$, $$B_2 = 0.50\,\text{T}$$, $$R_2 = 7\,\Omega$$

Using $$-(2)$$ for each coil and taking the ratio:

$$\frac{S_{V1}}{S_{V2}} = \frac{\dfrac{N_1A_1B_1}{kR_1}}{\dfrac{N_2A_2B_2}{kR_2}} = \frac{N_1A_1B_1R_2}{N_2A_2B_2R_1}$$

Substituting the given values:

$$N_1A_1B_1R_2 = 15 \times 3.6\times10^{-3} \times 0.25 \times 7$$
$$= 15 \times 0.0036 \times 0.25 \times 7$$
$$= 0.0945$$

$$N_2A_2B_2R_1 = 21 \times 1.8\times10^{-3} \times 0.50 \times 5$$
$$= 21 \times 0.0018 \times 0.50 \times 5$$
$$= 0.0945$$

Therefore

$$\frac{S_{V1}}{S_{V2}} = \frac{0.0945}{0.0945} = 1$$

Thus, the voltage sensitivities of $$M_1$$ and $$M_2$$ are equal, giving the ratio

$$S_{V1} : S_{V2} = 1 : 1$$

Option A is correct.

When a charged particle ($$q, m$$) enters a uniform magnetic field ($$B$$) with velocity ($$v$$) perpendicular to the field, it executes uniform circular motion.

Radius ($$r$$), $$r = \frac{mv}{qB}$$

Region 2:

The particle enters Region 2 with magnetic field $$B_2$$. It describes a semicircular arc and returns to the interface.

Radius in Region 2: $$r_2 = \frac{mv}{qB_2}$$

Displacement along the interface: $$d_2 = 2r_2 = \frac{2mv}{qB_2}$$

Assuming a positive charge moving downward, the magnetic force ($$\vec{F} = q\vec{v} \times \vec{B}$$) directed into the page causes the particle to curve to the right.

Region 1:

Upon returning to the interface, the particle enters Region 1 with magnetic field $$B_1$$. Because the velocity vector is now reversed (pointing upward), the force direction also reverses.

Radius in Region 1: $$r_1 = \frac{mv}{qB_1}$$

Displacement along the interface: $$d_1 = 2r_1 = \frac{2mv}{qB_1}$$

The particle now curves to the left.

Net displacement ($$S$$): $$S = |d_1 - d_2|$$

$$S = 2r_1 - 2r_2 = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$$

$$S = \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2$$

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

To find the magnetic field $$B$$ at a distance $$r$$ from the center of the wire, we use Ampere's Circuital Law:

$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} $$

Case 1: Inside the wire ($$r < a$$)

Since the current $$I$$ is uniformly distributed across the circular cross-section, the current enclosed ($$I_{enc}$$) by an Amperian loop of radius $$r$$ is proportional to the area:

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

Applying Ampere's Law:

$$ B (2\pi r) = \mu_0 \left( I \frac{r^2}{a^2} \right) $$

$$ B = \left( \frac{\mu_0 I}{2\pi a^2} \right) r $$

Therefore, inside the wire, the magnetic field increases linearly with distance: {$$B \propto r$$}.

Case 2: Outside the wire

For a point outside the wire, the Amperian loop encloses the entire current $$I$$:

$$ I_{enc} = I $$

Applying Ampere's Law:

$$ B (2\pi r) = \mu_0 I $$

$$ B = \frac{\mu_0 I}{2\pi r} $$

Therefore, outside the wire, the magnetic field decreases inversely with distance: {$$B \propto \frac{1}{r}$$}

$$\vec{B}_{\text{net}} = \vec{B}_{\text{top wire}} + \vec{B}_{\text{arc}} + \vec{B}_{\text{bottom wire}}$$

For top semi-infinite wire aligned with the center line: $$B_{\text{top wire}} = 0$$

For the circular arc subtending angle $$\theta = \frac{3\pi}{2}$$ into the page ($$\otimes$$):

$$B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \theta \implies B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right)$$

For bottom semi-infinite wire at perpendicular distance $$a$$ into the page ($$\otimes$$):

$$B_{\text{bottom wire}} = \frac{\mu_0 I}{4\pi a} (\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi a} (1)$$

Total field magnitude: $$B_{\text{net}} = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right) + \frac{\mu_0 I}{4\pi a} (1) \implies B_{\text{net}} = \frac{\mu_0 I}{4\pi a} \left[\frac{3\pi}{2} + 1\right]$$

The magnetic field inside a long, closely-wound solenoid is

$$B = \mu H$$

where $$\mu$$ is the permeability of the medium and $$H$$ is the magnetic field strength produced by the current.

In vacuum (or air) $$\mu = \mu_0$$, so the field is

$$B_0 = \mu_0 H$$ $$-(1)$$

When the core is filled by a material of magnetic susceptibility $$\chi$$, its permeability becomes

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

Therefore the new magnetic field is

$$B = \mu H = \mu_0(1 + \chi)H$$ $$-(2)$$

Divide $$(2)$$ by $$(1)$$ to find the ratio of the two fields:

$$\frac{B}{B_0} = \frac{\mu_0(1 + \chi)H}{\mu_0 H} = 1 + \chi$$

The fractional (i.e. relative) increase in the field is

$$\frac{B - B_0}{B_0} = (1 + \chi) - 1 = \chi$$

Hence the percentage increase is

$$\chi \times 100\%$$

For magnesium $$\chi_{mg} = 1.2 \times 10^{-5}$$, so

Percentage increase $$= 1.2 \times 10^{-5} \times 100\% = 1.2 \times 10^{-3}\%$$

Option A expresses this value as $$\dfrac{6}{5} \times 10^{-3}\% = 1.2 \times 10^{-3}\%$$, which matches our result.

Therefore, the correct choice is Option A.

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.

Maximum magnetic field occurs at the surface ($$r = a$$): $$B_{\text{max}} = \frac{\mu_0 I}{2\pi a}$$

Condition given: $$B = \frac{1}{2}B_{\text{max}} = \frac{\mu_0 I}{4\pi a}$$

Inside the wire ($$r_1 < a$$):

$$B_{\text{in}} = \frac{\mu_0 I r_1}{2\pi a^2}$$

$$\frac{\mu_0 I r_1}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \implies \frac{r_1}{a} = \frac{1}{2} \implies r_1 = \frac{a}{2}$$

Outside the wire ($$r_2 > a$$):

$$B_{\text{out}} = \frac{\mu_0 I}{2\pi r_2}$$

$$\frac{\mu_0 I}{2\pi r_2} = \frac{\mu_0 I}{4\pi a} \implies r_2 = 2a$$

Let the wire lie along the z-axis with current $$I \hat{k}$$.

The particle is in the x-z plane at distance $$r$$ from the wire.

$$B = \frac{\mu_0 I}{2\pi r} (-\hat{j})$$

Taking coordinates $$(r, z)$$ where $$r$$ is the perpendicular distance from the wire:

$$\vec{v} = v_r \hat{i} + v_z \hat{k}$$

$$\vec{F} = q(\vec{v} \times \vec{B}) = q \left[(v_r \hat{i} + v_z \hat{k}) \times \left(-\frac{\mu_0 I}{2\pi r} \hat{j}\right)\right] = -\frac{q\mu_0 I v_r}{2\pi r}\hat{k} + \frac{q\mu_0 I v_z}{2\pi r}\hat{i}$$

Equation of motion along the z-direction:

$$M \frac{dv_z}{dt} = -\frac{q\mu_0 I}{2\pi r} v_r \implies M \frac{dv_z}{dt} = -\frac{q\mu_0 I}{2\pi r} \frac{dr}{dt}$$

$$M dv_z = -\frac{q\mu_0 I}{2\pi} \frac{dr}{r}$$

Integrating from the initial state ($$r=a, v_z=v_0$$) to the turning point ($$r=x$$ where $$v_r=0$$):

Since magnetic force does no work, total speed remains $$v_0$$. At the turning point, $$v_r = 0 \implies v_z = -v_0$$.

$$\int_{v_0}^{-v_0} M dv_z = -\frac{q\mu_0 I}{2\pi} \int_{a}^{x} \frac{dr}{r}$$

$$-2M v_0 = -\frac{q\mu_0 I}{2\pi} \ln\left(\frac{x}{a}\right) \implies \ln\left(\frac{x}{a}\right) = -\frac{4\pi M v_0}{q\mu_0 I}$$

$$x = a e^{-\frac{4\pi M v_0}{q\mu_0 I}}$$

The magnetisation $$M$$ produced in a material kept in an external magnetic field $$H$$ is written as
$$M = \chi\,H$$, where $$\chi$$ is called the magnetic susceptibility.

The magnetic induction $$B$$ inside the same material is given by
$$B = \mu\,H$$, where $$\mu$$ is the magnetic permeability of the material.

For vacuum (or free space) the induction is $$B_0 = \mu_0\,H$$, so the relative permeability of the material is defined as
$$\mu_r = \frac{\mu}{\mu_0}$$.

By definition we can also write
$$B = \mu_0\,(H + M)$$. Substituting $$M = \chi\,H$$ gives
$$B = \mu_0\,(H + \chi H) = \mu_0\,(1 + \chi)\,H$$.

Comparing the two expressions for $$B$$:
$$\mu\,H = \mu_0\,(1 + \chi)\,H$$.
Cancelling the common factor $$H$$, we get
$$\mu = \mu_0\,(1 + \chi)$$.

Divide both sides by $$\mu_0$$:
$$\frac{\mu}{\mu_0} = 1 + \chi$$.

The left side is just $$\mu_r$$, so
$$\mu_r = 1 + \chi \quad$$(relation between $$\mu_r$$ and $$\chi$$).

Rearranging for susceptibility:
$$\chi = \mu_r - 1 = \frac{\mu}{\mu_0} - 1$$.

Thus the correct option is
$$\chi = \frac{\mu}{\mu_0} - 1$$ → Option A.

Resistance of $$ABC = r$$

Resistance of $$ADC = 2r$$

$$i_1 = I \times \frac{2r}{r + 2r} = \frac{2}{3}I$$

$$i_2 = I \times \frac{r}{r + 2r} = \frac{1}{3}I$$

$$B_0 = \frac{\mu_0 i}{4\pi \left(\frac{a}{2}\right)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a}\left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2}\mu_0 i}{2\pi a}$$

$$B_{ABC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_1}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{2}{3}I\right)}{\pi a} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} \quad (\odot)$$

$$B_{ADC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_2}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{1}{3}I\right)}{\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a} \quad (\otimes)$$

$$B_{\text{net}} = B_{ABC} - B_{ADC} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} - \frac{\sqrt{2}\mu_0 I}{3\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$$

For a magnetic field $$\mathbf{B}$$, Gauss’s law for magnetism states $$\nabla\!\cdot\!\mathbf{B}=0$$ $$-(1)$$.

Equation $$(1)$$ means the net magnetic flux through any closed surface is zero. Hence there are no sources or sinks of the magnetic field analogous to electric charges. In other words, isolated north poles or south poles (magnetic monopoles) are absent. This validates Assertion (A).

Because $$\nabla\!\cdot\!\mathbf{B}=0$$ everywhere, magnetic field lines can neither start nor end; they must be continuous and form closed loops. Therefore Reason (R) is also a correct statement.

Moreover, the continuity of magnetic field lines (closed loops) is the direct physical manifestation of $$\nabla\!\cdot\!\mathbf{B}=0$$, which is precisely why magnetic monopoles do not exist. Thus Reason (R) correctly explains Assertion (A).

So, both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

Answer: Option C

For a circular coil of radius $$R$$ carrying current $$I$$, the standard results for magnetic field are:

• At the centre of the coil (on its plane) $$B_1 = \frac{\mu_0 I}{2R} \quad -(1)$$

• On the axis of the coil at a distance $$x$$ from the centre $$B_2 = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \quad -(2)$$

To obtain the required ratio, divide $$(2)$$ by $$(1)$$:

$$\frac{B_2}{B_1} = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \times \frac{2R}{\mu_0 I} = \frac{R^{3}}{\left(R^{2}+x^{2}\right)^{3/2}} \quad -(3)$$

Rewrite $$(3)$$ more compactly:

$$\frac{B_2}{B_1} = \left(\frac{R^{2}}{R^{2}+x^{2}}\right)^{3/2} \quad -(4)$$

The problem specifies $$x : R = 3 : 4$$, so choose $$x = 3k$$ and $$R = 4k$$ for some constant $$k$$.

Substitute these into $$(4)$$:

$$\frac{B_2}{B_1} = \left( \frac{(4k)^{2}}{(4k)^{2} + (3k)^{2}} \right)^{3/2} = \left( \frac{16k^{2}}{16k^{2} + 9k^{2}} \right)^{3/2} = \left( \frac{16}{25} \right)^{3/2} \quad -(5)$$

Evaluate the power:

$$\left( \frac{16}{25} \right)^{3/2} = \frac{16^{3/2}}{25^{3/2}} = \frac{( \sqrt{16} )^{3}}{( \sqrt{25} )^{3}} = \frac{4^{3}}{5^{3}} = \frac{64}{125}$$

Hence, $$\displaystyle \frac{B_2}{B_1} = \frac{64}{125}$$, or in ratio form $$64 : 125$$.

Therefore, the correct option is Option C (64 : 125).

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

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

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

$$r = \frac{m}{qB} \sqrt{\frac{2E}{m}}$$

$$r = \frac{\sqrt{2mE}}{qB}$$

$$r \propto \sqrt{E}$$

Graph D correctly represents the $$r \propto \sqrt{E}$$ curve.

Assertion (A): An electron moves with constant velocity in a straight line in a uniform magnetic field.

The magnetic force on a moving charge is $$\vec{F} = q\vec{v} \times \vec{B}$$. If $$\vec{v} \parallel \vec{B}$$, then $$\vec{v} \times \vec{B} = 0$$ and the force is zero, so the electron moves in a straight line with constant velocity. TRUE.

Reason (R): The magnetic field is along the direction of velocity of the electron.

This is exactly the condition that makes the cross product zero. TRUE.

R correctly explains A: when $$\vec{B} \parallel \vec{v}$$, there is no magnetic force, so the electron travels in a straight line.

The correct answer is Option C: Both (A) and (R) are true and (R) is the correct explanation of (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.

The magnetic field inside a solenoid is given by:

$$ B = \mu_0 n I = \mu_0 \frac{N}{L} I $$

We are given that $$B = 2.9 \times 10^{-4}$$ T, $$N = 200$$, $$I = 0.29$$ A, and $$\mu_0 = 4\pi \times 10^{-7}$$ T·m/A.

We rearrange this formula to solve for $$L$$:

$$ L = \frac{\mu_0 N I}{B} = \frac{4\pi \times 10^{-7} \times 200 \times 0.29}{2.9 \times 10^{-4}} $$

It follows that

$$ L = \frac{4\pi \times 10^{-7} \times 58}{2.9 \times 10^{-4}} = \frac{4\pi \times 58 \times 10^{-7}}{2.9 \times 10^{-4}} $$

Simplifying further gives

$$ = \frac{4\pi \times 20 \times 10^{-7}}{10^{-4}} = 4\pi \times 20 \times 10^{-3} = 80\pi \times 10^{-3} \text{ m} $$

Finally,

$$ = 8\pi \text{ cm} $$

Therefore, the answer is 8.

The magnetic force $$F$$ on a straight conductor of length $$l$$ carrying current $$I$$ in a uniform magnetic field of magnitude $$B$$ is given by the formula

$$F = I\,l\,B\,\sin\theta$$,

where $$\theta$$ is the angle between the direction of current and the magnetic field.

Here the wire is placed perpendicular to the field, so $$\theta = 90^{\circ}$$ and $$\sin 90^{\circ} = 1$$. Thus,

$$F = I\,l\,B$$ $$-(1)$$

Convert the length from centimetres to metres:
$$l = 4.0\ \text{cm} = 4.0 \times 10^{-2}\ \text{m} = 0.04\ \text{m}$$

Substitute $$I = 8\ \text{A}$$, $$l = 0.04\ \text{m}$$ and $$B = 0.15\ \text{T}$$ into $$(1)$$:

$$F = 8 \times 0.04 \times 0.15$$

First multiply $$8 \times 0.04 = 0.32$$.
Then multiply $$0.32 \times 0.15 = 0.048$$.

So, $$F = 0.048\ \text{N}$$.

To express the force in milli-newtons (1 mN = $$10^{-3}$$ N):
$$0.048\ \text{N} = 0.048 \times 10^{3}\ \text{mN} = 48\ \text{mN}$$.

Therefore, the magnetic force acting on the wire is $$\mathbf{48\ mN}$$.

Straight segments $$AB$$ and $$CD$$: Their lines of extension pass through center $$O$$. $$\implies B_{AB} = 0 \quad \text{and} \quad B_{CD} = 0$$

Inner semicircular arc $$BC$$ ($$Radius = R_2$$): The current flows clockwise. According to the Right-Hand Rule, its magnetic field ($$B_2$$) points into the page ($$\otimes$$).

Outer semicircular arc $$DA$$ ($$Radius = R_1$$): The current flows counter-clockwise. According to the Right-Hand Rule, its magnetic field ($$B_1$$) points out of the page ($$\odot$$).

$$B_1 = \frac{\mu_0 I}{4 R_1} = \frac{(4\pi \times 10^{-7}) \times 12}{4 \times 6\pi} = \frac{48\pi \times 10^{-7}}{24\pi} = 2 \times 10^{-7}\text{ T} \quad (\odot)$$

$$B_2 = \frac{\mu_0 I}{4 R_2} = \frac{(4\pi \times 10^{-7}) \times 12}{4 \times 4\pi} = \frac{48\pi \times 10^{-7}}{16\pi} = 3 \times 10^{-7}\text{ T} \quad (\otimes)$$

$$B_{\text{net}} = B_2 - B_1 = (3 \times 10^{-7}) - (2 \times 10^{-7}) = 1 \times 10^{-7}\text{ T} \quad (\otimes)$$

$$k = 1$$

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

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

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

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

$$I_X = 5 \text{ A}$$

$$I_Y = 4 \text{ A}$$

Distance between wires X and Y is $$6 \text{ cm}$$

Point P is at a distance of $$4 \text{ cm}$$ to the right of wire Y.

The total distance of point P from wire X is:

$$ r_X = 6 \text{ cm} + 4 \text{ cm} = 10 \text{ cm} = 0.1 \text{ m} $$

The distance of point P from wire Y is:

$$ r_Y = 4 \text{ cm} = 0.04 \text{ m} $$

The magnetic field due to a long straight wire is given by the formula:

$$ B = \frac{\mu_0 I}{2\pi r} $$

Calculate the magnetic field at point P due to wire X ($$B_X$$):

$$ B_X = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1} $$

$$ B_X = \frac{2 \times 10^{-7} \times 5}{0.1} $$

$$ B_X = 10 \times 10^{-6} \text{ T} = 1 \times 10^{-5} \text{ T} $$

By the right-hand grip rule, with current flowing upwards, the direction of $$B_X$$ at point P is into the plane of the paper.

Calculate the magnetic field at point P due to wire Y ($$B_Y$$):

$$ B_Y = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.04} $$

$$ B_Y = \frac{2 \times 10^{-7} \times 4}{0.04} $$

$$ B_Y = 2 \times 10^{-5} \text{ T} $$

By the right-hand grip rule, with current flowing downwards, the direction of $$B_Y$$ at point P is out of the plane of the paper.

Since the magnetic fields are in opposite directions, the resultant magnetic field ($$B_{net}$$) is the difference between their magnitudes:

$$ B_{net} = B_Y - B_X $$

$$ B_{net} = 2 \times 10^{-5} \text{ T} - 1 \times 10^{-5} \text{ T} $$

$$ B_{net} = 1 \times 10^{-5} \text{ T} $$

Comparing this with the given expression $$x \times 10^{-5} \text{ T}$$, we get the value of $$x$$:

$$ x = 1 $$

A proton and a deuteron with the same kinetic energy enter a uniform magnetic field perpendicular to $$\vec{B}$$. We need the ratio of their radii of circular paths.

When a charged particle moves perpendicular to a magnetic field, it follows a circular path with radius:

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

Since $$KE = \frac{1}{2}mv^2$$, we get $$mv = \sqrt{2m \cdot KE}$$. Therefore:

$$ r = \frac{\sqrt{2m \cdot KE}}{qB} $$

For the proton: mass = $$m_p$$, charge = $$e$$, so $$r_p = \frac{\sqrt{2m_p \cdot KE}}{eB}$$

For the deuteron: mass = $$2m_p$$ (approximately twice the proton mass), charge = $$e$$ (same charge as proton), so:

$$ r_d = \frac{\sqrt{2 \cdot 2m_p \cdot KE}}{eB} = \frac{\sqrt{2} \cdot \sqrt{2m_p \cdot KE}}{eB} = \sqrt{2} \cdot r_p $$

The ratio is:

$$ \frac{r_d}{r_p} = \sqrt{2} : 1 $$

The correct answer is Option (1): $$\sqrt{2} : 1$$.

Galvanometer relationship: $$I = k\theta \implies I \propto \theta$$

Given initial conditions:

$$I_1 = 200\ \mu\text{A}$$

$$\theta_1 = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}\text{ rad}$$

Given final condition: $$\theta_2 = \frac{\pi}{10}\text{ rad}$$

$$\frac{I_2}{I_1} = \frac{\theta_2}{\theta_1}$$

$$\frac{I_2}{200} = \frac{\left(\frac{\pi}{10}\right)}{\left(\frac{\pi}{3}\right)} = \frac{3}{10} \implies I_2 = 200 \times \frac{3}{10} = 60\ \mu\text{A}$$

$$\text{For (A): Magnetic susceptibility }\chi_m\text{ of a material is independent of magnetising field }H\implies Y = \text{constant}\implies \text{(III)}$$

$$\text{For (B): Inside a current-carrying wire }(x < a)\text{, Ampere's Law gives }B = \frac{\mu_0 I x}{2\pi a^2}\implies Y \propto X\implies \text{(I)}$$

$$\text{For (C): Outside a current-carrying wire }(x > a)\text{, Ampere's Law gives }B = \frac{\mu_0 I}{2\pi x}\implies Y \propto \frac{1}{X}\implies \text{(IV)}$$

$$\text{For (D): Inside a finite solenoid, }B\text{ is uniform near the center and drops by half near the edges}\implies \text{(II)}$$

We need to find the ratio of the magnetic field at distances $$\frac{a}{2}$$ and $$2a$$ from the axis of a long straight wire of radius $$a$$ carrying a steady, uniformly distributed current $$I$$.

Recall that Ampere's law states: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$ where $$I_{enc}$$ is the current enclosed by the Amperian loop.

Inside the wire at a distance $$r$$ from the axis, the current is uniformly distributed over the cross-section of radius $$a$$. The enclosed current is proportional to the enclosed area: $$I_{enc} = I \cdot \frac{\pi r^2}{\pi a^2} = I \cdot \frac{r^2}{a^2}$$.

Applying Ampere's law with a circular Amperian loop of radius $$r$$ gives $$B \cdot 2\pi r = \mu_0 \cdot I \cdot \frac{r^2}{a^2}$$, so $$B_{inside} = \frac{\mu_0 I r}{2\pi a^2}$$. At $$r = \frac{a}{2}$$, this becomes $$B_1 = \frac{\mu_0 I \cdot \frac{a}{2}}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$$.

Outside the wire at a distance $$r$$ from the axis, the entire current $$I$$ is enclosed, so Ampere's law gives $$B \cdot 2\pi r = \mu_0 I$$ and hence $$B_{outside} = \frac{\mu_0 I}{2\pi r}$$. At $$r = 2a$$, this becomes $$B_2 = \frac{\mu_0 I}{2\pi \cdot 2a} = \frac{\mu_0 I}{4\pi a}$$.

The ratio of these fields is $$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1}$$, so $$B_1 : B_2 = 1 : 1$$.

The correct answer is Option (2): 1 : 1.

A proton moves with constant velocity, meaning the net force on it is zero.

(A) $$E = 0, B = 0$$: No force, proton moves at constant velocity. Possible.

(B) $$E = 0, B \neq 0$$: If the proton moves parallel to $$\vec{B}$$, the magnetic force $$q\vec{v} \times \vec{B} = 0$$. Possible.

(C) $$E \neq 0, B = 0$$: The electric force $$q\vec{E} \neq 0$$, so the proton would accelerate. Not possible.

(D) $$E \neq 0, B \neq 0$$: If $$q\vec{E} + q\vec{v} \times \vec{B} = 0$$, the electric and magnetic forces can cancel. Possible.

The correct options are (A), (B) and (D) only, which corresponds to Option (3).

Magnetic field $$B = 2 \times 10^{-3}$$ T along +Y. Rectangular loop 20 cm × 10 cm in Y-Z plane, current 5 A anticlockwise w.r.t. negative X-axis.

The area vector of the loop in Y-Z plane points along $$\pm\hat{x}$$. Since current is anticlockwise when viewed from negative X-axis, by the right-hand rule, the area vector points along $$-\hat{x}$$.

$$\vec{m} = I\vec{A} = 5 \times (20 \times 10 \times 10^{-4})(-\hat{x}) = 5 \times 0.02(-\hat{x}) = -0.1\hat{x}$$ A m$$^2$$.

Torque: $$\vec{\tau} = \vec{m} \times \vec{B} = (-0.1\hat{x}) \times (2 \times 10^{-3}\hat{y})$$

$$= -0.1 \times 2 \times 10^{-3}(\hat{x} \times \hat{y}) = -2 \times 10^{-4}\hat{z}$$

Magnitude: $$2 \times 10^{-4}$$ N m, direction: along negative Z-direction.

The correct answer is Option B: $$2 \times 10^{-4}$$ N m along negative Z-direction.

$$\vec{dl} = \Delta x \hat{i} = 10^{-2}\hat{i}\ \text{m}$$

$$\vec{r} = 0.5\hat{j}\ \text{m} \implies \hat{r} = \hat{j}$$

$$\vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\Delta x \hat{i} \times \hat{j})}{r^2}$$

$$dB = 10^{-7} \times \frac{10 \times 10^{-2} \times \sin(90^\circ)}{(0.5)^2}$$

$$dB = 10^{-7} \times \frac{10^{-1}}{0.25} = 4 \times 10^{-8}\ \text{T}$$

This question asks for the correct expressions for the electrostatic force and magnetic force acting on a charge $$q$$ moving with velocity $$\vec{v}$$.

Recall that the electrostatic force on a charge $$q$$ in an electric field $$\vec{E}$$ is given by:

$$\vec{F_1} = q\vec{E}$$

This force acts in the direction of $$\vec{E}$$ for a positive charge and opposite to $$\vec{E}$$ for a negative charge. Importantly, this force does not depend on the velocity of the charge.

The magnetic force on a charge $$q$$ moving with velocity $$\vec{v}$$ in a magnetic field $$\vec{B}$$ is given by:

$$\vec{F_2} = q(\vec{v} \times \vec{B})$$

Key features of this force:

- It involves the cross product of velocity and magnetic field, not a dot product.

- The order matters: it is $$\vec{v} \times \vec{B}$$, not $$\vec{B} \times \vec{v}$$ (they differ by a negative sign).

- This force is always perpendicular to both $$\vec{v}$$ and $$\vec{B}$$, so it does no work on the charge.

- If the charge is stationary ($$\vec{v} = 0$$), the magnetic force is zero.

Comparing with the given options shows that option (1) gives $$\vec{F_1} = q\vec{E}$$ and $$\vec{F_2} = q(\vec{V} \times \vec{B})$$, which matches both formulas correctly.

Option (2) incorrectly uses $$\vec{B}$$ for the electrostatic force.

Option (3) has the cross product in the wrong order ($$\vec{B} \times \vec{V}$$).

Option (4) incorrectly uses dot products instead of cross products.

The correct answer is Option (1): $$\vec{F_1} = q\vec{E}$$, $$\vec{F_2} = q(\vec{V} \times \vec{B})$$.

$$U = -MB \cos\theta$$

$$\text{Most stable position (parallel): } \theta_1 = 0^\circ$$

$$\text{Most unstable position (anti-parallel): } \theta_2 = 180^\circ$$

$$W = \Delta U = -MB(\cos\theta_2 - \cos\theta_1)$$

$$\implies W = -MB(\cos 180^\circ - \cos 0^\circ) = -MB(-1 - 1) = 2MB$$

$$\implies W = 2 \times 0.5 \times (8 \times 10^{-2}) = 8 \times 10^{-2}\text{ J}$$

$$B_1 = B_2 = \frac{\mu_0 I}{2a}$$

$$B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$$ ($$B_1$$ and $$B_2$$ are perpendicular vectors)

$$B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2} \implies B_{\text{net}} = \sqrt{2} \cdot \frac{\mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2}a}$$

We need to find the mass ratio $$\frac{m_X}{m_Y}$$ of two particles with equal charges accelerated through the same potential difference, making circular paths of radii $$R_1$$ and $$R_2$$.

When a particle of mass m and charge q is accelerated through a potential $$V$$, its kinetic energy satisfies $$\frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}}$$. In a magnetic field, the circular radius is given by $$R = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mV/q}}{B} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$, so that $$R \propto \sqrt{m}$$ when q, V, and B are constant.

Hence, $$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y}}$$ and thus $$\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$$.

Therefore, the answer is Option B.

$$\text{From the figure, the field is out of the page along the z-axis: } \vec{B} = B_0\hat{k}$$

$$\text{Immersed initial point (left wire entering field) to final point (right wire leaving field):}$$ $$\vec{L}_{\text{eff}} = 2R\hat{i}$$

$$\vec{F} = i(\vec{L}_{\text{eff}} \times \vec{B}) = i(2R\hat{i} \times B_0\hat{k})$$

$$\vec{F} = 2iRB_0(\hat{i} \times \hat{k}) = -2iRB_0\hat{j}$$

Statement I: "When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account."

This is true. For time-varying currents, the electromagnetic field itself carries momentum. The mechanical momentum of the charges alone is not conserved; one must include the electromagnetic field momentum (given by $$\frac{1}{c^2}\vec{E} \times \vec{B}$$ per unit volume) for Newton's third law (conservation of momentum) to hold. This is a fundamental result in classical electrodynamics.

Statement II: "Ampere's circuital law does not depend on Biot-Savart's law."

This is false. For steady (magnetostatic) currents, Ampere's circuital law can be derived from the Biot-Savart law. The Biot-Savart law gives the magnetic field due to a current element, and integrating it appropriately yields Ampere's law. They are mathematically equivalent descriptions of magnetostatics, so Ampere's law does depend on (and is derivable from) the Biot-Savart law.

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

Ampere's Circuital Law: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$$

For a circular Amperean loop drawn completely outside the cable (at a radius greater than the outer conductor's radius):

$$I_{\text{enclosed}} = I_{\text{central}} + I_{\text{outer}}$$

Since the two conductors carry equal currents in opposite directions:

$$I_{\text{enclosed}} = I + (-I) = 0$$

$$\implies \oint \vec{B} \cdot d\vec{l} = 0 \implies B = 0$$

Answer: Option (A): outside the cable

The z-axis carries an infinitely long straight wire with current $$I$$ along $$+z$$. At a point with position vector $$\vec{r}=x\,\hat{i}+y\,\hat{j}$$ in the $$xy$$-plane its magnetic field is

$$\vec{B}=\frac{\mu_0 I}{2\pi r}\,\hat{\phi} ,\qquad r=\sqrt{x^{2}+y^{2}}$$

For current in $$+z$$, $$\hat{\phi}= -\sin\phi\,\hat{i}+\cos\phi\,\hat{j}$$, where $$\tan\phi=\dfrac{y}{x}$$. With $$x$$ variable and constant $$y=a$$ (our path), we have

$$\sin\phi=\frac{a}{r},\qquad \cos\phi=\frac{x}{r}$$

Hence

$$\vec{B}= \frac{\mu_0 I}{2\pi r}\Bigl(-\frac{a}{r}\,\hat{i} +\frac{x}{r}\,\hat{j}\Bigr) =\frac{\mu_0 I}{2\pi r^{2}}\bigl(-a\,\hat{i}+x\,\hat{j}\bigr)$$

The required line integral is taken along the straight segment from $$P_1(-\sqrt{3}a,\,a,\,0)$$ to $$P_2(a,\,a,\,0)$$. Along this segment $$y=a,\;z=0$$ and $$x$$ varies, so

$$d\vec{l}=dx\,\hat{i}$$

Therefore

$$\vec{B}\cdot d\vec{l}= \frac{\mu_0 I}{2\pi r^{2}}\bigl(-a\,\hat{i}+x\,\hat{j}\bigr)\!\cdot\!(dx\,\hat{i}) =-\frac{\mu_0 I\,a}{2\pi (x^{2}+a^{2})}\,dx$$

Integrate from $$x=-\sqrt{3}a$$ to $$x=a$$:

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I\,a}{2\pi}\int_{-\sqrt{3}a}^{a}\frac{dx}{x^{2}+a^{2}}$$

Use the standard integral $$\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\!\Bigl(\frac{x}{a}\Bigr)$$:

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I}{2\pi}\Bigl[\tan^{-1}\!\Bigl(\frac{x}{a}\Bigr)\Bigr]_{x=-\sqrt{3}a}^{x=a}$$

Evaluate the limits:

$$\tan^{-1}(1)=\frac{\pi}{4},\qquad \tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}$$

Thus

$$\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l} =-\frac{\mu_0 I}{2\pi}\!\Bigl(\frac{\pi}{4}-\bigl(-\frac{\pi}{3}\bigr)\Bigr) =-\frac{\mu_0 I}{2\pi}\!\Bigl(\frac{7\pi}{12}\Bigr) =-\frac{7\mu_0 I}{24}$$

The negative sign only indicates that $$\vec{B}$$ and $$d\vec{l}$$ are oppositely directed. The question asks for the magnitude, hence

$$\Bigl|\int_{P_1}^{P_2}\vec{B}\cdot d\vec{l}\Bigr| =\frac{7\mu_0 I}{24}$$

Option A which is: $$\dfrac{7\mu_0 I}{24}$$

Using magnetic moment of a circular loop: $$M = I A = I(\pi r^2)$$

Using torque due to uniform vertical magnetic field $$B_0$$ when loop tilts by angle $$\theta$$:

$$\tau_m = \vert{}\vec{M} \times \vec{B}_0\vert{} = M B_0 \cos\theta = \pi r^2 I B_0 \cos\theta$$

Using restoring torque due to gravity acting at the center of mass:

$$\tau_g = m g (r \sin\theta) = m g r \sin\theta$$

Equating torques at rotational equilibrium about the axis $$PQ$$:

$$\tau_m = \tau_g \implies \pi r^2 I B_0 \cos\theta = m g r \sin\theta$$

$$\frac{\sin\theta}{\cos\theta} = \frac{\pi r^2 I B_0}{m g r} \implies \tan\theta = \frac{\pi r I B_0}{m g}$$

Using kinetic energy from potential difference: $$K = qV$$

Using path radius in a magnetic field: $$R = \frac{\sqrt{2mK}}{qB_0} = \frac{1}{B_0}\sqrt{\frac{2mV}{q}}$$

$$R = \frac{1}{0.1}\sqrt{\frac{2 \times A_M \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}} = 10\sqrt{A_M \times 400 \times 10^{-8}} = 0.02\sqrt{A_M}\text{ m} = 2\sqrt{A_M}\text{ cm}$$

Using geometry of the semicircular path back to the detector plane:

$$x = 2R = 4\sqrt{A_M}\text{ cm}$$

Evaluating $$x$$ for $$H^+$$ ion ($$A_M = 1$$): $$x = 4\sqrt{1} = 4\text{ cm}$$

Evaluating $$x$$ for $$A_M = 144$$: $$x = 4\sqrt{144} = 4 \times 12 = 48\text{ cm}$$

Evaluating detector limits for $$1 \le A_M \le 196$$:

$$x_{\text{min}} = 4\sqrt{1} = 4\text{ cm} \implies x_0 = 4\text{ cm}$$

$$x_{\text{max}} = 4\sqrt{196} = 4 \times 14 = 56\text{ cm} \implies x_1 = 56\text{ cm}$$

$$\text{Height } (x_1 - x_0) = 56 - 4 = 52\text{ cm}$$

Minimum field width $$w$$ to complete the semicircle for $$A_M = 196$$:

$$R_{\text{max}} = 2\sqrt{196} = 28\text{ cm} \implies w_{\text{min}} = R_{\text{max}} = 28\text{ cm}$$

Answer: Option (A), Option (B)

A charge $$q = 4.0\;\mu\text{C} = 4.0 \times 10^{-6}$$ C moves with velocity $$\vec{v} = 4.0 \times 10^6\;\hat{j}$$ m/s in a magnetic field $$\vec{B} = 2\hat{k}$$ T. The magnetic force on a moving charge is given by $$\vec{F} = q(\vec{v} \times \vec{B}).$$

To compute the cross product $$\vec{v} \times \vec{B},$$ observe that $$\vec{v} \times \vec{B} = (4.0 \times 10^6\;\hat{j}) \times (2\hat{k}) = 4.0 \times 10^6 \times 2\;(\hat{j} \times \hat{k}).$$ Using the rule $$\hat{j} \times \hat{k} = \hat{i},$$ it follows that $$\vec{v} \times \vec{B} = 8.0 \times 10^6\;\hat{i}.$$

Multiplying by the charge yields $$\vec{F} = q(\vec{v} \times \vec{B}) = 4.0 \times 10^{-6} \times 8.0 \times 10^6\;\hat{i},$$ so $$\vec{F} = 32\;\hat{i}\text{ N}.$$

The answer is $$x = 32$$.

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.

We need to find the work done in turning a current-carrying coil through 90° in a uniform magnetic field. First, recall that the magnetic moment of a coil is $$M = NIA$$, where $$N$$ is the number of turns, $$I$$ is the current, and $$A$$ is the area. The potential energy of a magnetic dipole in a field is $$U = -MB\cos\theta$$, and the work done is the change in potential energy, given by $$W = U_f - U_i$$.

Using the specified values, the magnetic moment is $$M = NIA = 100 \times 1 \times 10^{-3} \times 5 \times 10^{-3}$$, which simplifies to $$M = 100 \times 5 \times 10^{-6} = 5 \times 10^{-4}\,\text{A·m}^2$$.

The plane of the coil is initially perpendicular to the magnetic field, so the magnetic moment, which is normal to the coil’s plane, is parallel to the field. Therefore, $$\theta_i = 0°$$. After rotating through 90°, the angle becomes $$\theta_f = 90°$$.

Applying the expression for work, we have $$W = U_f - U_i = (-MB\cos 90°) - (-MB\cos 0°) = 0 - (-MB) = MB$$.

Substituting $$M = 5 \times 10^{-4}\,\text{A·m}^2$$ and $$B = 0.20\,\text{T}$$ yields $$W = 5 \times 10^{-4} \times 0.20 = 1 \times 10^{-4}\,\text{J} = 100\,\mu\text{J}$$.

The correct answer is 100 $$\mu$$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.

We need to find the magnetic field at the centroid of an equilateral triangle carrying current $$I = 2$$ A with side $$a = 4\sqrt{3}$$ cm.

The perpendicular distance from the centroid to each side of an equilateral triangle is:

$$d = \frac{a}{2\sqrt{3}} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \text{ cm} = 0.02 \text{ m}$$

Now, each side subtends an angle of $$60°$$ on either end as seen from the centroid. The magnetic field due to a finite straight conductor at perpendicular distance $$d$$ is:

$$B_{\text{one side}} = \frac{\mu_0 I}{4\pi d}(\sin\alpha + \sin\beta)$$

For each side, $$\alpha = \beta = 60°$$, so:

$$B_{\text{one side}} = \frac{\mu_0 I}{4\pi d} \times 2\sin 60° = \frac{\mu_0 I}{4\pi d} \times \sqrt{3}$$

By symmetry, all three sides produce magnetic fields in the same direction (perpendicular to the plane, by the right-hand rule). Hence the total field is:

$$B = 3 \times \frac{\mu_0 I}{4\pi d} \times \sqrt{3} = \frac{3\sqrt{3}\mu_0 I}{4\pi d}$$

Substituting values:

$$B = \frac{3\sqrt{3} \times 10^{-7} \times 2}{0.02} = \frac{6\sqrt{3} \times 10^{-7}}{0.02} = 3\sqrt{3} \times 10^{-5} \text{ T}$$

So, the answer is $$3\sqrt{3} \times 10^{-5}$$ T.

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%.

The magnetic force per unit length, $$f = \frac{\mu_0 i_1 i_2}{2\pi d}$$

Current in wire $$PQ$$ ($$i_1$$) $$= \frac{I_{total}}{2}$$

Current in wire $$RS$$ ($$i_2$$) $$= \frac{I_{total}}{2}$$

Inside the wire for the region $$r < a$$, the current enclosed by an Amperian loop of radius $$r$$ depends on the cross-sectional area:

$$I_{\text{enclosed}} = I \cdot \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2}$$

Applying Ampere's Law for this internal region: $$\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$$

$$B \cdot (2\pi r) = \mu_0 I \frac{r^2}{a^2}$$ $$\implies$$ $$B = \frac{\mu_0 I r}{2\pi a^2}$$

This shows that $$B \propto r$$ inside the wire.

Outside the wire for the region $$r > a$$, the Amperian loop of radius $$r$$ encloses the total current: $$I_{\text{enclosed}} = I$$

Applying Ampere's Law for this external region: $$\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$$

$$B \cdot (2\pi r) = \mu_0 I$$ $$\implies$$ $$B = \frac{\mu_0 I}{2\pi r}$$

This shows that $$B \propto \frac{1}{r}$$ outside the wire.

The circular loop lies in the $$yz$$ plane and its axis is along the $$x$$-axis.

Current is anticlockwise when viewed from $$+z$$-direction.

We need variation of :

$$B_y(0,a,z)$$

where point lies in the $$yz$$ plane at fixed :

$$y=a$$

with

$$a<R$$

Due to symmetry :

For every point at $$+z$$ there exists a symmetric point at $$-z$$.

Hence,

$$B_y(0,a,z)$$

is an even function of $$z$$.

At centre :

$$z=0$$

the magnetic field has maximum $$y$$-component.

As $$|z|$$ increases, the field decreases symmetrically.

Therefore :

- $$B_y$$ is maximum at $$z=0$$
- decreases symmetrically on both sides
- remains positive throughout

Hence the correct graph is the symmetric bell-shaped curve.

Final Answer :

$$\boxed{\text{Option D}}$$

Solution :

For a circular arc carrying current, magnetic field at centre is :

$$B = \frac{\mu_0 I \theta}{4\pi r}$$

For semicircle :

$$\theta = \pi$$

So,

$$B_{semi} = \frac{\mu_0 I}{4r}$$

Magnetic field due to straight wire at distance r :

$$B = \frac{\mu_0 I}{4\pi r}$$

Case A :

Magnetic field due to full circular loop :

$$B_1 = \frac{\mu_0 I}{2r}$$

Field due to straight wire :

$$B_2 = \frac{\mu_0 I}{2\pi r}$$

Both are in same direction.

$$B_0 = \frac{\mu_0 I}{2\pi r}(\pi+1)$$

Hence,

A → IV

Case B :

Magnetic field due to semicircle :

$$B_1 = \frac{\mu_0 I}{4r}$$

Fields due to two straight wires add :

$$B_2 = \frac{\mu_0 I}{2\pi r}$$

Hence,

$$B_0 = \frac{\mu_0 I}{4\pi r}(\pi+2)$$

Therefore,

B → I

Case C :

Magnetic field due to semicircle :

$$B_1 = \frac{\mu_0 I}{4r}$$

Field due to straight wire is opposite in direction.

Hence,

$$B_0 = \frac{\mu_0 I}{2\pi r}(\pi-1)$$

Therefore,

C → III

Case D :

Only semicircular arc contributes significantly.

$$B_0 = \frac{\mu_0 I}{4r}$$

Therefore,

D → II

Final Answer :

A-IV, B-I, C-III, D-II

Solution :

Magnetic field at the centre of a circular coil is given by :

$$B = \frac{\mu_0 NI}{2R}$$

Given initial number of turns :

$$N_1 = 4$$

Initial magnetic field :

$$B_1 = 32\text{ T}$$

When the wire is rewound into a single turn coil :

$$N_2 = 1$$

Since length of wire remains constant :

$$2\pi R_1N_1 = 2\pi R_2N_2$$

$$R_2 = 4R_1$$

Now,

$$B_2 = \frac{\mu_0 N_2 I}{2R_2}$$

$$= \frac{\mu_0 I}{2(4R_1)}$$

$$= \frac{\mu_0 I}{8R_1}$$

Initial field :

$$B_1 = \frac{\mu_0 (4)I}{2R_1}$$

$$= \frac{2\mu_0 I}{R_1}$$

Therefore,

$$\frac{B_2}{B_1} = \frac{\mu_0 I/8R_1}{2\mu_0 I/R_1}$$

$$= \frac{1}{16}$$

Hence,

$$B_2 = \frac{32}{16}$$

$$= 2\text{ T}$$

Final Answer :

$$2\text{ T}$$

The force between two parallel wires carrying currents is given by $$F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$. For the first wire, the current is $$I = 10$$ A and the separation is $$d = 5$$ cm, while for the second wire the current is $$I' = 20$$ A and the separation is $$d' = 2.5$$ cm.

The ratio of the forces is found by substituting these values into the formula for force, which yields $$\frac{F_2}{F_1} = \frac{I'^2}{I^2} \times \frac{d}{d'} = \frac{400}{100} \times \frac{5}{2.5} = 4 \times 2 = 8$$, and hence $$F_2 = 8F_1$$.

Two long parallel wires $$P$$ and $$Q$$ carry equal current $$I = 10$$ A, separated by distance $$d = 5$$ cm. We need to find the new force when the distance is halved and the currents are doubled. The magnetic force per unit length between two parallel wires is $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$ and for a length $$L = 10$$ cm $$= 0.1$$ m of wire $$P$$ this becomes $$F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$.

For the original configuration, substituting $$I_1 = I_2 = 10$$ A and $$d = 0.05$$ m gives $$F_1 = \frac{\mu_0 \times 10 \times 10 \times 0.1}{2\pi \times 0.05} = \frac{10\mu_0}{\pi \times 0.1} = \frac{100\mu_0}{\pi}.$$

In the new configuration where the distance is halved ($$d' = 2.5$$ cm) and currents are doubled ($$I' = 20$$ A), we have $$F_2 = \frac{\mu_0 \times 20 \times 20 \times 0.1}{2\pi \times 0.025} = \frac{40\mu_0}{\pi \times 0.05} = \frac{800\mu_0}{\pi}.$$

The ratio of the forces is $$\frac{F_2}{F_1} = \frac{(I')^2 / d'}{I^2 / d} = \frac{(20)^2 \times 0.05}{(10)^2 \times 0.025} = \frac{400 \times 0.05}{100 \times 0.025} = \frac{20}{2.5} = 8.$$ Alternatively, doubling each current increases the force by a factor of $$2 \times 2 = 4$$ and halving the distance increases it by a factor of $$2$$, giving a total factor of $$4 \times 2 = 8$$. Hence $$F_2 = 8F_1.$$

Answer: Option (1): $$\boxed{8F_1}$$.

The work done in rotating a magnetic dipole in a uniform magnetic field is:

$$ W = -mB(\cos\theta_2 - \cos\theta_1) $$

where $$m = 5.0$$ A m$$^2$$, $$B = 0.4$$ T.

Initial position: parallel ($$\theta_1 = 0°$$)

Final position: antiparallel ($$\theta_2 = 180°$$)

$$ W = -mB(\cos 180° - \cos 0°) = -5.0 \times 0.4 \times (-1 - 1) $$

$$ W = -2.0 \times (-2) = 4 \text{ J} $$

A circular loop of radius R carries a current i. The magnetic field at the centre of the loop is given by $$B_{\text{centre}} = \frac{\mu_0 i}{2R}$$. At a point on the axis of the loop at a distance $$x$$ from the centre, the field has the form $$B_{\text{axis}} = \frac{\mu_0 i R^2}{2(R^2 + x^2)^{3/2}}$$.

When this point is located at a distance equal to the loop radius (that is, when $$x = R$$), the expression simplifies as follows:
$$B_{\text{axis}} = \frac{\mu_0 i R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 i R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 i R^2}{2 \cdot 2\sqrt{2} \cdot R^3} = \frac{\mu_0 i}{4\sqrt{2}R}$$.

The ratio of the field at the centre to the field on the axis at $$x = R$$ is therefore
$$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\mu_0 i/(2R)}{\mu_0 i/(4\sqrt{2}R)} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\,. $$ This gives a final result of $$2\sqrt{2} : 1$$.

We need to find the ratio of the magnetic field at the centre of a circular loop to the field at a point on the axis.

The magnetic field at the centre is given by $$B_{\text{centre}} = \frac{\mu_0 i}{2R}$$, and the field on the axis at a distance $$x$$ from the centre is $$B_{\text{axis}} = \frac{\mu_0 i R^2}{2(R^2 + x^2)^{3/2}}$$.

The ratio is $$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{(R^2 + x^2)^{3/2}}{R^3}$$, and at $$x = R$$ this becomes $$\frac{(2R^2)^{3/2}}{R^3} = \frac{2\sqrt{2}R^3}{R^3} = 2\sqrt{2}$$, so the ratio is $$2\sqrt{2}:1$$.

This matches Option C: $$2\sqrt{2}:1$$.

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.

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

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

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

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

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

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

Find the magnetic intensity at the centre of a solenoid with 1200 turns, length 2 m, diameter 0.2 m, carrying a current of 2 A.

Calculate the number of turns per unit length.

$$n = \frac{N}{L} = \frac{1200}{2} = 600 \text{ turns/m}$$

Apply the formula for magnetic intensity (H).

The magnetic intensity (magnetic field strength H) inside a solenoid is:

$$H = nI = 600 \times 2 = 1200 \text{ A m}^{-1} = 1.2 \times 10^3 \text{ A m}^{-1}$$

Note: The diameter of the solenoid does not affect the magnetic intensity H inside it.

The correct answer is $$1.2 \times 10^3$$ A m$$^{-1}$$.

An electron has negative charge $$q = -e$$. The magnetic (Lorentz) force on a moving charge is given by
$$\mathbf{F}=q\,(\mathbf{v}\times\mathbf{B})$$

Velocity of the electron: $$\mathbf{v}=v\,\hat{i}$$ (along the +x-axis).
Magnetic field applied: $$\mathbf{B}=-B\,\hat{k}$$ (along the −z-axis).

First evaluate the cross product for a positive charge:
$$\mathbf{v}\times\mathbf{B}=v\hat{i}\times(-B\hat{k})$$
$$= -vB(\hat{i}\times\hat{k})$$

Using the right-hand rule for unit vectors, $$\hat{i}\times\hat{k}=-\hat{j}$$. Hence
$$\mathbf{v}\times\mathbf{B}=-vB(-\hat{j})=vB\,\hat{j}$$ (towards +y).

For an electron, multiply by $$q=-e$$:
$$\mathbf{F}=-e(vB\,\hat{j})=-evB\,\hat{j}$$

Therefore the magnetic force is along the negative y-axis. Statement B is correct, while statement A is incorrect.

The velocity is perpendicular to the magnetic field ( $$\mathbf{v}\perp\mathbf{B}$$ ), so the electron undergoes uniform circular motion in the plane perpendicular to $$\mathbf{B}$$ with constant speed and radius $$r=\dfrac{mv}{eB}$$. Hence statement E is also correct.

Statements C (no force) and D (continues along +x) are wrong.

The correct set of statements is B and E only → Option C.

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

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

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

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

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

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

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.

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

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

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

Solution :

Magnetic field due to a semi-infinite straight current carrying wire at perpendicular distance \(r\) is :

$$B = \frac{\mu_0 I}{4\pi r}$$

At point \(O\), there are four semi-infinite wires each at distance :

$$r = a$$

Using right hand thumb rule, magnetic field due to each wire at \(O\) is directed into the plane of paper.

Therefore, fields add together.

Magnetic field due to one wire :

$$B_1 = \frac{\mu_0 I}{4\pi a}$$

Total magnetic field :

$$B = 4B_1$$

$$= 4\left(\frac{\mu_0 I}{4\pi a}\right)$$

$$= \frac{\mu_0 I}{\pi a}$$

Final Answer :

$$\frac{\mu_0 I}{\pi a}$$

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 area of each turn of a moving coil galvanometer.

Deflection $$\theta = 0.05$$ rad, current $$I = 10$$ mA = 0.01 A, torsional constant $$C = 4.0 \times 10^{-5}$$ N m/rad, magnetic field $$B = 0.01$$ T, number of turns $$N = 200$$.

We start by noting that

In a moving coil galvanometer, the magnetic torque on the coil ($$\tau_m = NBIA$$) is balanced by the restoring torque of the suspension wire ($$\tau_r = C\theta$$):

$$ NBIA = C\theta $$

where $$A$$ is the area of each turn.

Next,

$$ A = \frac{C\theta}{NBI} $$

$$ A = \frac{4.0 \times 10^{-5} \times 0.05}{200 \times 0.01 \times 0.01} $$

$$ A = \frac{2.0 \times 10^{-6}}{0.02} = 1.0 \times 10^{-4}\;\text{m}^2 $$

Converting to cm$$^2$$: $$1.0 \times 10^{-4}$$ m$$^2$$ = $$1.0$$ cm$$^2$$.

The area of each turn is 1.0 cm$$^2$$.

The correct answer is Option 2: 1.0.

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.

We need to find the condition for equal magnetic moments of two circular coils A and B.

Magnetic moment of a coil:

$$M = NIA = NI\pi r^2$$

For coil A: $$M_A = N_A I_A \pi r_A^2 = N_A I_A \pi (0.1)^2 = 0.01\pi N_A I_A$$

For coil B: $$M_B = N_B I_B \pi r_B^2 = N_B I_B \pi (0.2)^2 = 0.04\pi N_B I_B$$

Setting $$M_A = M_B$$:

$$0.01\pi N_A I_A = 0.04\pi N_B I_B$$

$$N_A I_A = 4 N_B I_B$$

The correct answer is Option 3: $$N_A I_A = 4N_B I_B$$.

We are given that a charged particle moving in a magnetic field $$\vec{B}$$ has velocity components both along $$\vec{B}$$ and perpendicular to $$\vec{B}$$.

Decompose the velocity. Let the velocity be split as $$\vec{v} = v_{\parallel}\hat{B} + v_{\perp}\hat{n}$$, where $$v_{\parallel}$$ is along $$\vec{B}$$ and $$v_{\perp}$$ is perpendicular to $$\vec{B}$$.

Effect of the perpendicular component. The magnetic force $$\vec{F} = q(\vec{v} \times \vec{B})$$ acts only on $$v_{\perp}$$. Since this force is always perpendicular to $$v_{\perp}$$, it causes the particle to move in a circular path in the plane perpendicular to $$\vec{B}$$.

Effect of the parallel component. Since $$\vec{v}_{\parallel} \times \vec{B} = 0$$, there is no magnetic force along $$\vec{B}$$. The particle moves with constant velocity $$v_{\parallel}$$ along the field direction.

Combined motion. The superposition of circular motion (perpendicular to $$\vec{B}$$) and uniform linear motion (along $$\vec{B}$$) gives a helical (spiral) path with its axis along the direction of $$\vec{B}$$.

The correct answer is Option B: helical path with the axis along magnetic field B.

We have a charge $$q = 2 \mu C = 2 \times 10^{-6}$$ C accelerated through a potential difference $$V = 100$$ V, entering a magnetic field $$B = 4$$ mT $$= 4 \times 10^{-3}$$ T where it traces a semicircle of radius $$r = 3$$ cm $$= 0.03$$ m.

The kinetic energy gained by the charge after acceleration is

Now, the radius of circular motion in a magnetic field gives us

So $$mv = qBr$$. Substituting into the kinetic energy equation,

Solving for mass,

Hence, the mass of the charged particle is $$144 \times 10^{-18}$$ kg. So, the answer is $$144$$.

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

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

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

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

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

Magnetic force on a current carrying conductor is given by:

F=BILsin⁡θ

where
$$B=0.75T$$
$$I=2A$$
L= length of the conductor
θ=angle between current direction and magnetic field.

The magnetic field is parallel to the current in the 13 cm side.
Hence, the magnetic field is along the hypotenuse of the right triangle.

We need the force on the 5 cm side.

In the $$5\text{-}12\text{-}13$$ right triangle, the angle between the 5 cm side and the 13 cm side is obtained using:

$$\sin\theta=\frac{12}{13}$$

Length of the 5 cm side:

$$L=5cm=\frac{5}{100}m$$

Now,

$$F=0.75\times2\times\frac{5}{100}\times\frac{12}{13}$$

$$F=\frac{90}{1300}$$

$$F=\frac{9}{130}N$$

Given,

$$F=\frac{x}{130}N$$

Therefore,

$$x=9$$

For the wire to be in equilibrium with zero tension in the supporting leads, the magnetic force must balance gravity.

$$BIL = mg$$

Given: $$m = 40$$ g $$= 0.04$$ kg, $$L = 50$$ cm $$= 0.5$$ m, $$B = 0.40$$ T, $$g = 10$$ m/s².

$$0.40 \times I \times 0.5 = 0.04 \times 10$$

$$0.2I = 0.4$$

$$I = 2 \text{ A}$$

The current required is $$\mathbf{2}$$ A.

We need to find the magnetic field at the centre O of a semicircular arc carrying a current of 14 A with radius 2.2 cm.

We begin with the Biot-Savart law, which gives the magnetic field at the centre of a circular arc subtending an angle $$\theta$$ (in radians) at the centre: $$B = \frac{\mu_0 I \theta}{4\pi r}$$ where $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A is the permeability of free space, $$I$$ is the current, and $$r$$ is the radius.

For a semicircular arc, $$\theta = \pi$$ radians. Substituting into the formula yields $$B = \frac{\mu_0 I \pi}{4\pi r} = \frac{\mu_0 I}{4r}$$.

Substituting the given values $$I = 14$$ A, $$r = 2.2$$ cm $$= 0.022$$ m, and $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A into this expression gives $$B = \frac{4\pi \times 10^{-7} \times 14}{4 \times 0.022}$$.

To simplify, the numerator is $$4\pi \times 10^{-7} \times 14 = 56\pi \times 10^{-7}$$ and the denominator is $$4 \times 0.022 = 0.088$$. This gives $$B = \frac{56\pi \times 10^{-7}}{0.088} = \frac{56\pi}{0.088} \times 10^{-7}$$.

Next, $$\frac{56}{0.088} = \frac{56000}{88} = \frac{7000}{11} \approx 636.36$$ so $$B \approx 636.36 \times \pi \times 10^{-7} \approx 636.36 \times 3.14 \times 10^{-7}$$ and hence $$B \approx 1998.2 \times 10^{-7} = 2.0 \times 10^{-4} \text{ T}$$.

Converting to the required units confirms that $$B = 2.0 \times 10^{-4}$$ T = $$2 \times 10^{-4}$$ T. Therefore, the magnetic field at the centre of the semicircular arc is 2 $$\times 10^{-4}$$ T.

$$I = \frac{q}{T} = \frac{ev}{2\pi r}$$

$$B = \frac{\mu_0 I}{2r}$$

$$B = \frac{\mu_0 e v}{4\pi r^2}$$

$$B = \frac{(4\pi \times 10^{-7}) \cdot (1.6 \times 10^{-19}) \cdot (6.76 \times 10^6)}{4\pi \cdot (0.52 \times 10^{-10})^2}$$

$$B = 40 \text{ T}$$

For a circular arc, the field at the center is $$B = \frac{\mu_0 I}{4\pi R} \theta$$, where $$\theta$$ is the angle in radians. A semicircle ($$\theta = \pi$$) produces $$B = \frac{\mu_0 I}{4R}$$

For the straight parts of the wire, the point $$O$$ lies exactly on the line of the conductor. This means the angle ($$\theta$$) between the current element $$d\vec{l}$$ and the position vector $$\vec{r}$$ is $$0^\circ$$ or $$180^\circ$$. Since $$\sin(0^\circ) = \sin(180^\circ) = 0$$, the magnetic field contribution from both straight sections is zero.

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

$$B = \frac{12\pi \times 10^{-7}}{\frac{4\pi}{10}} = \frac{12\pi \times 10^{-7} \times 10}{4\pi}$$

$$B = 3 \times 10 \times 10^{-7} = 3 \times 10^{-6}\text{ T}$$ 

$$\mathbf{B = 3\text{ }\mu\text{T}}$$

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.

The magnetic field at the centre of a current-carrying coil of radius $$r$$ is:

$$B_{\text{centre}} = \frac{\mu_0 I}{2r}$$

The magnetic field at a distance $$r$$ from the centre on the axis of the coil is:

$$B_{\text{axis}} = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 \cdot 2\sqrt{2} \cdot r^3} = \frac{\mu_0 I}{4\sqrt{2} \cdot r}$$

Now taking the ratio:

$$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2} r}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} = \sqrt{8}$$

We are given that the ratio is $$\sqrt{x} : 1$$, so:

$$\sqrt{x} = 2\sqrt{2} = \sqrt{8}$$

$$x = 8$$

Hence, the answer is $$8$$.

We have two identical circular loops of radius $$R = 20\;\text{cm} = 0.2\;\text{m}$$, carrying current $$I = \sqrt{2}\;\text{A}$$, placed in perpendicular planes.

The magnetic field at the centre of a circular loop is

Each loop produces a magnetic field at the common centre:

Since the two loops are in perpendicular planes, their magnetic fields are perpendicular to each other. So the net magnetic field is

Hence, the net magnetic field at the centre is $$628 \times 10^{-8}$$ T. So, the answer is $$628$$.

We need to find the pitch of the helical path taken by a proton with kinetic energy 2.0 eV in a magnetic field of magnitude $$\frac{\pi}{2} \times 10^{-3}$$ T, where the angle between the velocity and the magnetic field is 60°.

The kinetic energy relation $$KE = \frac{1}{2}mv^2$$ allows us to solve for the proton’s speed as $$v = \sqrt{\frac{2 \cdot KE}{m}}$$; substituting $$KE = 2.0 \times 1.6 \times 10^{-19}\,\text{J}$$ and $$m = 1.6 \times 10^{-27}\,\text{kg}$$ gives $$v = \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}} = \sqrt{4 \times 10^{8}} = 2 \times 10^4\,\text{m/s}$$.

The component of this velocity parallel to the magnetic field is $$v_{\parallel} = v\cos60^\circ = 2 \times 10^4 \times \tfrac12 = 10^4\,\text{m/s}$$.

The proton’s circular motion in the plane perpendicular to the field has period $$T = \frac{2\pi m}{qB}$$, and with $$m = 1.6 \times 10^{-27}\,\text{kg}$$, $$q = 1.6 \times 10^{-19}\,\text{C}$$, and $$B = \frac{\pi}{2} \times 10^{-3}\,\text{T}$$ one finds $$T = \frac{2\pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \frac{\pi}{2} \times 10^{-3}} = \frac{2\pi \times 1.6 \times 10^{-27}}{0.8\pi \times 10^{-22}} = \frac{2 \times 1.6}{0.8} \times 10^{-5} = 4 \times 10^{-5}\,\text{s}$$.

The pitch of the helix follows from multiplying the parallel speed by the period as $$\text{Pitch} = v_{\parallel} \times T = 10^4 \times 4 \times 10^{-5} = 0.4\,\text{m} = 40\,\text{cm}$$, so that the result is $$\boxed{40\ \text{cm}}$$.

Solution :

Magnetic field due to a long straight current carrying wire is :

$$B = \frac{\mu_0 I}{2\pi r}$$

Given :

Current in first wire,

$$I_1 = 8\text{ A}$$

Current in second wire,

$$I_2 = 15\text{ A}$$

Distance between wires,

$$d = 7\text{ cm}$$

Since point P is equidistant from both wires and the lines joining P to the wires are perpendicular to each other, triangle formed is right angled isosceles.

Hence, distance of P from each wire :

$$r = \frac{7}{\sqrt{2}}\text{ cm}$$

$$= \frac{7}{1.4}\text{ cm}$$

$$= 5\text{ cm} = 0.05\text{ m}$$

Magnetic field due to first wire :

$$B_1 = \frac{\mu_0 I_1}{2\pi r}$$

$$= \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.05}$$

$$= 32 \times 10^{-6}\text{ T}$$

Magnetic field due to second wire :

$$B_2 = \frac{\mu_0 I_2}{2\pi r}$$

$$= \frac{4\pi \times 10^{-7} \times 15}{2\pi \times 0.05}$$

$$= 60 \times 10^{-6}\text{ T}$$

Since the magnetic fields are perpendicular,

$$B = \sqrt{B_1^2 + B_2^2}$$

$$= \sqrt{32^2 + 60^2}\times 10^{-6}$$

$$= \sqrt{4624}\times 10^{-6}$$

$$= 68 \times 10^{-6}\text{ T}$$

Final Answer :

$$68$$

We need to find how the magnetic field $$B$$ varies with distance $$r$$ from the centre of a long straight wire of radius $$R$$ carrying a uniformly distributed current $$I$$, for $$r < R$$.

By applying Ampere’s circuital law to a circular Amperian loop of radius $$r$$ inside the wire, one writes $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$$. Because the current is uniformly distributed over the cross section, the enclosed current is $$I_{enc} = I \frac{\pi r^2}{\pi R^2} = I \frac{r^2}{R^2}$$. By symmetry, $$B$$ is tangential and constant along the loop, so $$B\,(2\pi r) = \mu_0 I \frac{r^2}{R^2}$$, which yields $$B = \frac{\mu_0 I r}{2\pi R^2}$$.

Hence, inside the wire, the magnetic field varies as $$B \propto r$$. The correct answer is Option A: $$B \propto r$$.

Statement I: Susceptibilities of paramagnetic and ferromagnetic substances increase with decrease in temperature.

For paramagnetic materials, Curie's law states $$\chi = \frac{C}{T}$$, so susceptibility increases as temperature decreases. This is true.

For ferromagnetic materials, below the Curie temperature, the susceptibility increases significantly as temperature decreases (the material becomes more strongly magnetized). Above the Curie temperature, it follows the Curie-Weiss law $$\chi = \frac{C}{T - T_c}$$, which also increases as temperature decreases toward $$T_c$$. So Statement I is true.

Statement II: Diamagnetism is a result of orbital motions of electrons developing magnetic moments opposite to the applied magnetic field.

This is true. Diamagnetism arises because the applied magnetic field modifies the orbital motion of electrons, inducing a net magnetic moment that opposes the applied field (Lenz's law at the atomic level). This effect is present in all materials but is very weak.

Since both statements are true:

Hence, the correct answer is Option A.

Two charged particles with the same kinetic energy pass through a uniform magnetic field perpendicular to their motion. The ratio of radii is $$6:5$$, and the ratio of masses is $$9:4$$. We need to find the ratio of charges.

The radius of the circular path in a magnetic field is given by $$r = \frac{mv}{qB}$$, and since kinetic energy $$KE = \frac{1}{2}mv^2$$, we have $$v = \sqrt{\frac{2KE}{m}}$$.

Substituting this expression for velocity into the formula for the radius yields $$r = \frac{m}{qB}\sqrt{\frac{2KE}{m}} = \frac{\sqrt{2mKE}}{qB}$$.

Because both particles have the same kinetic energy and experience the same magnetic field, the ratio of their radii simplifies to $$\frac{r_1}{r_2} = \frac{\sqrt{m_1}/q_1}{\sqrt{m_2}/q_2} = \frac{q_2\sqrt{m_1}}{q_1\sqrt{m_2}}$$.

Using the given ratios $$\frac{r_1}{r_2} = \frac{6}{5}$$ and $$\frac{m_1}{m_2} = \frac{9}{4}$$, we substitute to obtain $$\frac{6}{5} = \frac{q_2}{q_1} \times \sqrt{\frac{9}{4}} = \frac{q_2}{q_1} \times \frac{3}{2}$$, which leads to $$\frac{q_2}{q_1} = \frac{6}{5} \times \frac{2}{3} = \frac{4}{5}$$.

It follows that $$\frac{q_1}{q_2} = \frac{5}{4}$$.

Hence, the correct answer is Option B.

A charged particle moves in a uniform magnetic field $$\vec{B} = (2\hat{i} + 3\hat{j}) \text{ T}$$. Its acceleration is $$\vec{a} = (\alpha\hat{i} - 4\hat{j}) \text{ m/s}^2$$. We need to find $$\alpha$$.

The magnetic force on a charged particle is given by:

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

Since $$\vec{F} = m\vec{a}$$, the acceleration $$\vec{a}$$ is proportional to $$\vec{v} \times \vec{B}$$.

The magnetic force is always perpendicular to the magnetic field. Therefore, the acceleration must also be perpendicular to $$\vec{B}$$:

$$\vec{a} \cdot \vec{B} = 0$$

$$(\alpha\hat{i} - 4\hat{j}) \cdot (2\hat{i} + 3\hat{j}) = 0$$

$$2\alpha + (-4)(3) = 0$$

$$2\alpha - 12 = 0$$

$$\alpha = 6$$

The correct answer is Option B: $$6$$.

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: $$\sqrt{3} : 4$$.

The magnetic field inside a long solenoid is given by:

$$B = \mu_0 n I$$

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

Initially, the magnetic field is:

$$B = \mu_0 n I$$

When the current is doubled ($$I' = 2I$$) and the number of turns per cm is halved ($$n' = \frac{n}{2}$$):

$$B' = \mu_0 n' I' = \mu_0 \times \frac{n}{2} \times 2I = \mu_0 n I = B$$

The new magnetic field equals the original magnetic field.

Hence, the correct answer is Option D.

We need to find the actual angle of dip ($$\delta$$) given the apparent angle of dip ($$\delta'$$) in a plane inclined to the magnetic meridian.

1. Identify the Given Values

• Angle of inclination with the magnetic meridian ($$\alpha$$) = $$45^\circ$$
• Apparent angle of dip ($$\delta'$$) = $$60^\circ$$

2. Formula for Apparent Dip

The relationship between the actual angle of dip ($$\delta$$), the apparent angle of dip ($$\delta'$$), and the angle with the magnetic meridian ($$\alpha$$) is given by the formula:

$$\tan \delta' = \frac{\tan \delta}{\cos \alpha}$$

Rearranging the formula to solve for the actual dip ($$\tan \delta$$):

$$\tan \delta = \tan \delta' \cdot \cos \alpha$$

3. Calculation

Substitute the known trigonometric values ($$\tan 60^\circ = \sqrt{3}$$ and $$\cos 45^\circ = \frac{1}{\sqrt{2}}$$) into the equation:

$$\tan \delta = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$$

Taking the inverse tangent to isolate $$\delta$$:

$$\delta = \tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$$