JEE Magnetism & Magnetic Materials Questions

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.

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.

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

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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.

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 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$$^2$$.

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.

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 ($$\theta_2 = 180°$$)

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

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

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.

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.

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.

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.

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

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

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

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

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

We need to find the relation between the magnetic moment and the orbital angular momentum of an electron revolving around a nucleus.

An electron moving in a circular orbit constitutes a current loop whose magnetic moment is given by $$\mu_L = iA$$, where $$i$$ is the current and $$A$$ is the area of the orbit.

The current due to the electron moving with period $$T$$ is $$i = \frac{e}{T} = \frac{ev}{2\pi r}$$, and the area of the circular orbit is $$A = \pi r^2$$.

Substituting these expressions gives $$\mu_L = \frac{ev}{2\pi r} \times \pi r^2 = \frac{evr}{2}$$.

Since the orbital angular momentum is $$L = mvr$$, it follows that $$vr = \frac{L}{m}$$, and therefore $$\mu_L = \frac{eL}{2m}$$.

For a negatively charged electron, the magnetic moment is opposite in direction to the angular momentum, so $$\vec{\mu}_L = -\frac{e\vec{L}}{2m}$$.

Hence, the correct answer is Option B.

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

Write the magnetic field with and without the magnetic material.

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

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

Calculate the fractional increase in magnetic field.

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

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

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

The correct answer is Option A.

We need to find the work done in rotating a bar magnet from 0° to 60° in a uniform magnetic field.

Magnetic moment: $$M = 2.0 \times 10^5$$ J T$$^{-1}$$

Magnetic field: $$B = 14 \times 10^{-5}$$ T

Initial angle: $$\theta_1 = 0°$$ (along the field)

Final angle: $$\theta_2 = 60°$$

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

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

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

$$W = 2.0 \times 10^5 \times 14 \times 10^{-5} \times (\cos 0° - \cos 60°)$$

$$W = 28 \times \left(1 - \frac{1}{2}\right)$$

$$W = 28 \times \frac{1}{2} = 14 \text{ J}$$

Hence, the correct answer is Option A.

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

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

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

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

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

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

The relationship between permeability and susceptibility is:

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

Substituting the values:

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

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

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

The correct answer is Option A.

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

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

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

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

Hence, the correct answer is Option 4.

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

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

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

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

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

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

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

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

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

The correct answer is Option C.

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

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

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

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

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

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

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

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

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

Statement I states that ferromagnetic property depends on temperature, and above a certain temperature (the Curie temperature), a ferromagnet becomes a paramagnet. This is correct — it is the well-established Curie temperature phenomenon.

Statement II states that at high temperature, the domain wall area of a ferromagnetic substance increases. This is incorrect. At high temperature, the domains actually break down or become disordered. The thermal energy disrupts the alignment of magnetic moments within domains. Above the Curie temperature, domain structure itself ceases to exist as the material becomes paramagnetic — it is not that domain walls increase in area. In fact, as temperature increases, thermal agitation destroys the long-range magnetic order, causing the domains to shrink and eventually disappear rather than domain walls expanding.

Therefore, Statement I is true but Statement II is false.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The numerical data are

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

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

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

divide by $$2\pi$$:

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

Now square both sides:

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

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

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

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

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

Calculate the numerator:

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

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

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

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

Substituting into the expression for $$B$$:

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

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

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

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

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

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

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

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

$$H = nI,$$

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

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

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

Magnetic susceptibility and relative permeability are connected through the familiar formula

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

Hence, for the first core material we have

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

and for the second core material we have

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

The magnetisations therefore become

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

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

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

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

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

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

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

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

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

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

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

Hence, the correct answer is Option 250.

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

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

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

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

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

The actual permeability inside the material is therefore

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

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

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

The increase in the field strength is

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

The fractional (or relative) increase is then

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

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

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

For aluminium we are given

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

Hence

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

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

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

Now we solve for $$x$$:

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

So, the answer is $$22$$.

We have a thin circular loop of radius $$a$$, carrying a current $$I$$ and possessing mass $$m$$. In a uniform magnetic field $$B$$ directed perpendicular to the plane of the loop, the loop behaves like a magnetic dipole. Its magnetic dipole moment is, by definition,

$$\mu = I \,A = I\,(\pi a^{2}).$$

Next, we need the moment of inertia of the loop about the diameter about which it is allowed to oscillate. For a thin circular ring, the perpendicular-axis theorem gives

$$I_{x}+I_{y}=I_{z}.$$

Here $$I_{z}=m a^{2}$$ (axis through the centre and perpendicular to the plane). Because the two diameters in the plane are equivalent, $$I_{x}=I_{y}$$; hence

$$2I_{x}=m a^{2}\quad\Longrightarrow\quad I_{x}=I_{y}=\frac{1}{2}m a^{2}.$$

Thus the moment of inertia about the diameter of oscillation is

$$I_{\text{m.o.i.}}=\frac{1}{2}m a^{2}.$$

When the loop is given a small angular displacement $$\theta$$ from its equilibrium position, the restoring magnetic torque is

$$\tau = \mu B \sin\theta.$$

For small angles, $$\sin\theta \approx \theta$$, so

$$\tau \approx \mu B\,\theta.$$

The rotational form of Newton’s second law states

$$I_{\text{m.o.i.}}\;\frac{d^{2}\theta}{dt^{2}} = -\tau,$$

the negative sign indicating that the torque is restoring. Substituting the expressions for $$\tau$$ and $$I_{\text{m.o.i.}}$$, we obtain

$$\frac{1}{2}m a^{2}\;\frac{d^{2}\theta}{dt^{2}} \;=\; -\,\mu B\,\theta.$$

Simplifying,

$$\frac{d^{2}\theta}{dt^{2}} + \underbrace{\left(\frac{2\mu B}{m a^{2}}\right)}_{\omega^{2}}\theta = 0.$$

This is the standard differential equation of simple harmonic motion, $$\displaystyle \frac{d^{2}\theta}{dt^{2}}+\omega^{2}\theta = 0,$$ whose angular frequency is

$$\omega = \sqrt{\frac{2\mu B}{m a^{2}}}.$$

The time period is related to angular frequency by the formula

$$T = \frac{2\pi}{\omega}.$$

Substituting $$\omega$$,

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\mu B}}.$$

Now we replace $$\mu$$ by $$I\pi a^{2}$$:

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\,I\pi a^{2} B}} = 2\pi\sqrt{\frac{m}{2I\pi B}}.$$

Notice that $$a^{2}$$ cancels out. Extracting the common factors inside the square root,

$$T = 2\pi\;\frac{\sqrt{m}}{\sqrt{2I\pi B}} = \frac{2\sqrt{\pi}\,\sqrt{m}}{\sqrt{2I B}} = \sqrt{\frac{2\pi m}{I B}}.$$

Hence, the correct answer is Option C.

We are told that a bar magnet placed in a uniform magnetic field experiences a torque. For a magnetic dipole, the relation between the torque $$\tau$$, the magnetic dipole moment $$M$$, the magnetic field $$B$$ and the angle $$\theta$$ between $$\vec M$$ and $$\vec B$$ is stated first:

$$\tau \;=\; M\,B\,\sin\theta$$

The numerical values given are

$$\tau = 0.018\ \text{N m}, \qquad B = 0.06\ \text{T}, \qquad \theta = 30^\circ.$$

We substitute these values into the formula to isolate $$M$$. We begin by writing

$$M = \dfrac{\tau}{B\,\sin\theta}.$$

Now, since $$\sin 30^\circ = \dfrac12,$$ we have

$$B\,\sin\theta \;=\; 0.06 \times \dfrac12 \;=\; 0.03.$$

So

$$M = \dfrac{0.018}{0.03} = 0.6\ \text{A m}^2.$$

Next we calculate the work needed to turn the magnet from its stable to its unstable equilibrium position. The potential energy $$U$$ of a magnetic dipole in a field is

$$U = -M\,B\,\cos\theta.$$

The stable equilibrium corresponds to $$\theta = 0^\circ$$ (magnet aligned with the field), and the unstable equilibrium corresponds to $$\theta = 180^\circ$$ (magnet anti-aligned). We compute the energies at these two orientations.

At $$\theta = 0^\circ$$:

$$U_{\text{stable}} = -M\,B\,\cos 0^\circ = -M\,B \times 1 = -M\,B.$$

At $$\theta = 180^\circ$$:

$$U_{\text{unstable}} = -M\,B\,\cos 180^\circ = -M\,B \times (-1) = +M\,B.$$

The work required is the increase in potential energy while going from the stable to the unstable state:

$$W = U_{\text{unstable}} - U_{\text{stable}} = (M\,B) - (-M\,B) = 2\,M\,B.$$

Substituting the values $$M = 0.6\ \text{A m}^2$$ and $$B = 0.06\ \text{T}$$ we get

$$W = 2 \times 0.6 \times 0.06 = 2 \times 0.036 = 0.072\ \text{J}.$$

Writing this in scientific notation,

$$W = 7.2 \times 10^{-2}\ \text{J}.$$

Hence, the correct answer is Option C.

We are asked to find the magnetic dipole moment of the iron rod after it is inserted in the solenoid and an electric current is sent through the turns. Let us introduce the symbols that will be used:

$$$V = 10^{-3}\,\text{m}^3 \qquad (\text{volume of the rod})$$$ $$$\mu_r = 1000 \qquad (\text{relative permeability of iron})$$$ $$$n = 10\,\text{turns cm}^{-1} \qquad (\text{number of turns per centimetre})$$$ $$$I = 0.5\,\text{A} \qquad (\text{current through the solenoid})$$$

First we convert the turn density into SI units (turns per metre). Since $$1\,\text{cm} = 10^{-2}\,\text{m},$$ we obtain

$$$n = 10\,\text{turns cm}^{-1} = 10 \times \frac{1}{10^{-2}}\,\text{turns m}^{-1} = 10 \times 100\,\text{turns m}^{-1} = 1000\,\text{turns m}^{-1}.$$$

Now we recall the definition of the magnetising field $$H$$ inside an ideal long solenoid. The standard formula is

$$H = n I,$$

where $$n$$ is the number of turns per metre and $$I$$ is the current in amperes. Substituting the known values we get

$$$H = (1000\,\text{turns m}^{-1})(0.5\,\text{A}) = 500\,\text{A m}^{-1}.$$$

Next we connect the magnetising field $$H$$ to the magnetisation $$M$$ of the iron core. For a linear magnetic material we use the relation

$$M = \chi_m H,$$

where $$\chi_m$$ is the magnetic susceptibility given by

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

For the present rod,

$$\chi_m = 1000 - 1 = 999.$$

Hence the magnetisation becomes

$$$M = 999 \times 500\,\text{A m}^{-1} = 499{,}500\,\text{A m}^{-1} = 4.995 \times 10^{5}\,\text{A m}^{-1}.$$$

The total magnetic dipole moment $$m$$ of the rod is obtained by multiplying the magnetisation $$M$$ by the volume $$V$$ (because magnetisation is magnetic moment per unit volume):

$$m = M V.$$

Substituting $$M = 4.995 \times 10^{5}\,\text{A m}^{-1}$$ and $$V = 10^{-3}\,\text{m}^3$$ gives

$$$m = (4.995 \times 10^{5}\,\text{A m}^{-1})(10^{-3}\,\text{m}^3) = 4.995 \times 10^{2}\,\text{A m}^2.$$$

In scientific notation this is essentially

$$m \approx 5 \times 10^{2}\,\text{A m}^2.$$

This numerical value exactly matches option B among the given choices.

Hence, the correct answer is Option B.

First, let us recall what the two keywords in the options mean.

Retentivity (also called remanence) is the residual value of magnetic induction $$B$$ left in the material when the magnetising field $$H$$ is reduced to zero. In simpler words, a material with large retentivity can hold on to its magnetism for a long time after the external field is removed.

Coercivity is the reverse magnetic field that must be applied to bring that residual induction back to zero. Thus a material with large coercivity needs a strong opposite field to demagnetise it; in other words, it strongly resists any attempt to lose its magnetisation.

Now we analyse the requirements for the two devices mentioned in the question.

For a permanent magnet, we want the magnet to remain magnetised under all ordinary circumstances. Therefore we require

$$\text{Retentivity : large}$$

so that a large residual magnetism is present, and

$$\text{Coercivity : large}$$

so that stray fields, heating, or mechanical shocks cannot easily demagnetise it.

For a transformer core, on the other hand, the magnetic field inside the core is made to reverse direction many times every second (50 Hz or 60 Hz). We need a material whose magnetisation follows the external field easily; otherwise energy will be wasted as heat. Hence, for a transformer we actually desire small coercivity and small retentivity, giving a narrow hysteresis loop and low hysteresis loss.

Scanning the given options, we see that Option D states

$$P : \text{large retentivity, large coercivity}$$

which is exactly the pair of properties we have just concluded to be essential for a permanent magnet. None of the other options simultaneously satisfies both conditions for the permanent magnet, and the statements given for the transformer in the other options are also not the ideal combination.

Hence, the correct answer is Option D.

We first recall the relation between the restoring torque acting on a magnetic dipole and the angular acceleration it produces. A magnetic dipole of moment $$\mu$$ kept in a uniform magnetic field $$B$$ experiences a torque

$$\tau = \mu B\sin\theta,$$

where $$\theta$$ is the angle between $$\vec\mu$$ and $$\vec B$$. For very small oscillations, $$\sin\theta \approx \theta$$, so the torque becomes

$$\tau \;=\; -\,\mu B\,\theta.$$

(The minus sign only indicates that the torque is restoring.)

In rotational dynamics the torque is also related to the angular acceleration $$\alpha$$ by

$$\tau = I\alpha = I\frac{d^{2}\theta}{dt^{2}},$$

where $$I$$ is the moment of inertia about the axis of rotation. Combining the two expressions, we obtain

$$I\frac{d^{2}\theta}{dt^{2}} = -\,\mu B\,\theta.$$

This is the differential equation of simple harmonic motion, with the square of the angular frequency given by

$$\omega^{2} = \frac{\mu B}{I}.$$

Hence the time-period of small oscillations is

$$T = 2\pi\sqrt{\frac{I}{\mu B}}.$$

Now we analyse the two given bodies one by one.

Hoop (ring)
We are told that the mass of the hoop is $$M$$ and the radius is $$R$$. About its own axis the moment of inertia is well known:

$$I_h = MR^{2}.$$

The magnetic moment of the hoop is twice that of the solid cylinder, so if we denote the cylinder’s magnetic moment by $$\mu_c$$, then

$$\mu_h = 2\mu_c.$$

Substituting these values in the period formula, we find

$$T_h = 2\pi\sqrt{\frac{I_h}{\mu_h B}} = 2\pi\sqrt{\frac{MR^{2}}{(2\mu_c)B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Solid cylinder
For a solid cylinder of the same mass $$M$$ and radius $$R$$, the moment of inertia about its own axis is

$$I_c = \frac{1}{2}MR^{2}.$$

Its magnetic moment is simply $$\mu_c.$$
Putting these quantities into the period formula gives

$$T_c = 2\pi\sqrt{\frac{I_c}{\mu_c B}} = 2\pi\sqrt{\frac{\dfrac{1}{2}MR^{2}}{\mu_c B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

We see that the square-root expressions for $$T_h$$ and $$T_c$$ are identical:

$$T_h = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}} \quad \text{and} \quad T_c = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Therefore

$$T_h = T_c.$$

Hence, the correct answer is Option B.

For a paramagnetic substance we use Curie’s law, which states that the magnetic susceptibility $$\chi$$ varies inversely with the absolute temperature $$T$$:

$$\chi = \dfrac{C}{T},$$

where $$C$$ is Curie’s constant (a material‐specific constant).

Because the same specimen is observed at two different temperatures, the value of $$C$$ remains unchanged. Thus we may write for the two sets of observations

$$\chi_1 T_1 = C = \chi_2 T_2.$$

Rearranging to obtain the unknown susceptibility $$\chi_2$$ at temperature $$T_2$$ we get

$$\chi_2 = \chi_1 \dfrac{T_1}{T_2}.$$

The data provided are

Initial susceptibility: $$\chi_1 = 2.8 \times 10^{-4}.$$

Initial temperature: $$T_1 = 350 \text{ K}.$$

Required temperature: $$T_2 = 300 \text{ K}.$$

Substituting these values into the rearranged Curie’s law expression:

$$\chi_2 = \left(2.8 \times 10^{-4}\right)\dfrac{350}{300}.$$

First evaluate the fraction $$\dfrac{350}{300}$$:

$$\dfrac{350}{300} = \dfrac{35}{30} = \dfrac{7}{6} = 1.166\overline{6}.$$

Now multiply this factor with $$2.8 \times 10^{-4}$$:

$$\chi_2 = 2.8 \times 10^{-4} \times 1.166\overline{6}.$$

Carrying out the multiplication step by step:

$$2.8 \times 1.166\overline{6} = 3.266\overline{6}.$$

Hence

$$\chi_2 \approx 3.266\overline{6} \times 10^{-4}.$$

Rounding to three significant figures,

$$\chi_2 \approx 3.267 \times 10^{-4}.$$

This value matches the entry in Option C.

Hence, the correct answer is Option C.

We begin by recalling that the magnetic moment of a uniformly magnetised body is related to its magnetisation by

$$\vec M \;=\; \frac{\text{total magnetic moment}}{\text{volume of the body}}\;.$$

Here the substance is a cube whose edge length is given as $$1\ \text{cm}$$. Converting this length into SI units, we have

$$1\ \text{cm}=1\times10^{-2}\ \text{m}=0.01\ \text{m}\,.$$

Now the volume of a cube is the cube of its side, so

$$V=(0.01\ \text{m})^{3}=0.01^{3}\ \text{m}^{3}=1\times10^{-6}\ \text{m}^{3}\,.$$

The total magnetic dipole moment of the cube is given as $$20\times10^{-6}\ \text{J/T}$$. Remembering that $$1\ \text{J/T}=1\ \text{A·m}^{2}$$, we can write

$$\mu =20\times10^{-6}\ \text{A·m}^{2}\,.$$

Substituting the values of $$\mu$$ and $$V$$ in the definition of magnetisation, we get

$$M=\frac{\mu}{V}=\frac{20\times10^{-6}\ \text{A·m}^{2}}{1\times10^{-6}\ \text{m}^{3}}=20\ \text{A/m}\,.$$

Next, magnetic susceptibility $$\chi_{m}$$ is defined as the ratio of magnetisation $$M$$ to the applied magnetic intensity $$H$$, that is,

$$\chi_{m}=\frac{M}{H}\,.$$

The applied magnetic intensity is given as $$60\times10^{3}\ \text{A/m}=60000\ \text{A/m}$$. Hence, substituting the known values,

$$\chi_{m}=\frac{20\ \text{A/m}}{60000\ \text{A/m}} =\frac{20}{60000} =\frac{2}{6000} =\frac{1}{3000} =3.33\times10^{-4}\,.$$

So we obtain

$$\chi_{m}\;\approx\;3.3\times10^{-4}\,.$$

Hence, the correct answer is Option D.

We have a magnetic dipole whose magnetic moment vector is given as $$\vec m = 10^{-2}\,\hat i\;{\rm A\,m^2}.$$

The external magnetic field is time-dependent and is specified by

$$\vec B(t)=B\cos\omega t\;\hat i,$$

where the numerical values are $$B = 1\ {\rm T}$$ and $$\omega = 0.125\ {\rm rad\,s^{-1}}.$$

At the instant $$t = 1\ {\rm s}$$ the field becomes

$$\vec B(1)=1\cos(0.125\times1)\;\hat i =\cos(0.125)\;\hat i\ {\rm T}.$$

The angle $$0.125$$ rad is small, so

$$\cos(0.125)\approx 0.9922.$$

Hence the field magnitude at this instant is approximately

$$B_1 = 0.9922\ {\rm T}.$$

The formula for the potential energy of a magnetic dipole in a magnetic field is

$$U = -\vec m\cdot\vec B.$$

Initially the dipole is parallel to the field, so the angle between $$\vec m$$ and $$\vec B$$ is $$0^\circ$$. Therefore

$$U_i = -mB_1\cos 0^\circ = -mB_1.$$

To reverse the dipole, we make it point opposite to the field; the angle becomes $$180^\circ$$. Then

$$U_f = -mB_1\cos 180^\circ = -mB_1(-1) = +mB_1.$$

The work done by an external agent equals the increase in potential energy:

$$W = U_f - U_i = \left(+mB_1\right) - \left(-mB_1\right) = 2mB_1.$$

Substituting $$m = 10^{-2}\ {\rm A\,m^2}$$ and $$B_1 = 0.9922\ {\rm T}$$, we get

$$W = 2 \times 10^{-2} \times 0.9922 = 0.019844\ {\rm J} \approx 0.02\ {\rm J}.$$

Hence, the correct answer is Option B.

We begin by recalling the theory of the small oscillations of a magnetic (compass) needle. When the needle is displaced slightly in the horizontal plane, the restoring torque is $$\tau = M B_H \sin\theta \approx M B_H \theta$$ for small $$\theta,$$ where $$M$$ is the magnetic moment and $$B_H$$ is the horizontal component of the earth’s magnetic field. For a body executing small-angle simple harmonic motion, the angular frequency is given by the standard formula

$$\omega = \sqrt{\dfrac{\text{restoring torque per unit angle}}{\text{moment of inertia}}} = \sqrt{\dfrac{M B_H}{I}}.$$

The ordinary (linear) frequency is $$f = \dfrac{\omega}{2\pi},$$ so we have

$$f = \dfrac{1}{2\pi}\sqrt{\dfrac{M}{I}}\,\sqrt{B_H}.$$

Everything except $$B_H$$ is constant for the same needle; hence

$$f \propto \sqrt{B_H} \quad\Longrightarrow\quad B_H \propto f^2.$$

Now we convert the given “oscillations per minute” to “oscillations per second” (hertz):

At the first place, the needle makes 30 oscillations in 60 s, so

$$f_1 = \dfrac{30}{60} = 0.50\ \text{Hz}.$$

At the second place, the needle makes 40 oscillations in 60 s, so

$$f_2 = \dfrac{40}{60} = 0.667\ \text{Hz}.$$

Using $$B_H \propto f^2,$$ we obtain

$$\dfrac{B_{H1}}{B_{H2}} = \left(\dfrac{f_1}{f_2}\right)^2 = \left(\dfrac{0.50}{0.667}\right)^2 = \left(0.75\right)^2 = 0.5625.$$

However, the question asks for the ratio of the total earth’s magnetic fields $$B_1$$ and $$B_2,$$ not just their horizontal components. The total field is related to its horizontal component by the well-known relation

$$B_H = B\cos\delta,$$

where $$\delta$$ is the dip (angle of inclination). Thus

$$B = \dfrac{B_H}{\cos\delta}.$$

So, for the two places,

$$\dfrac{B_1}{B_2} = \dfrac{B_{H1}/\cos\delta_1}{B_{H2}/\cos\delta_2} = \dfrac{B_{H1}}{B_{H2}}\;\dfrac{\cos\delta_2}{\cos\delta_1}.$$

The dips are $$\delta_1 = 45^\circ$$ and $$\delta_2 = 30^\circ.$$ We substitute their cosines:

$$\cos45^\circ = \dfrac{1}{\sqrt2} \approx 0.7071,\qquad \cos30^\circ = \dfrac{\sqrt3}{2} \approx 0.8660.$$

Therefore,

$$\dfrac{\cos\delta_2}{\cos\delta_1} = \dfrac{0.8660}{0.7071} \approx 1.2247.$$

Multiplying this with the earlier ratio of horizontal components, we get

$$\dfrac{B_1}{B_2} = 0.5625 \times 1.2247 \approx 0.688.$$

This value is very close to 0.7. Among the given options, the one that matches is 0.7.

Hence, the correct answer is Option D.

We have a bar magnet placed completely inside a current-carrying solenoid. In such a situation the magnetic field acting on the magnet is exactly the field produced by the solenoid itself. To make the magnet lose all its magnetisation, this external field must be equal in magnitude (but opposite in sense) to the magnet’s coercive field. Therefore the coercivity $$H_c$$ of the bar magnet is numerically the same as the magnetising field intensity $$H$$ created by the solenoid.

First we recall the standard formula for the field inside a long solenoid. For a solenoid having $$N$$ turns, length $$\ell$$, and carrying a current $$I$$,

$$B = \mu_0 \, n \, I,$$

where $$n$$ is the number of turns per unit length, $$n = \dfrac{N}{\ell}$$, and $$\mu_0$$ is the permeability of free space. The magnetic field intensity $$H$$ is related to the flux-density $$B$$ by

$$H = \dfrac{B}{\mu_0}.$$

Substituting $$B = \mu_0 n I$$ into this relation gives

$$H = \dfrac{\mu_0 n I}{\mu_0} = n I.$$

Thus, inside a long solenoid,

$$H = n I.$$

Now we substitute the numerical data given in the question. The solenoid has

$$N = 100 \text{ turns}, \qquad \ell = 0.2 \text{ m}.$$

So the turn density is

$$n = \dfrac{N}{\ell} = \dfrac{100}{0.2} = 500 \text{ turns per metre}.$$

The current flowing is

$$I = 5.2 \text{ A}.$$

Therefore the field intensity generated is

$$H = n I = 500 \times 5.2 = 2600 \text{ A/m}.$$

Since this field exactly cancels the magnet’s residual magnetism, the coercivity of the bar magnet is

$$H_c = 2600 \text{ A/m}.$$

Hence, the correct answer is Option 2.

We have the horizontal component of earth’s magnetic field at the place as $$B_h = 18 \times 10^{-6}\ \text{T}$$.

The magnetic needle, whose north and south poles have equal pole strength $$m = 1.8\ \text{A m}$$ and are separated by a length $$L = 0.12\ \text{m}$$, is suspended from its mid-point. In the earth’s field it comes to rest making an angle of $$45^{\circ}$$ with the horizontal.

The inclination (dip) angle of the earth’s field is defined by the relation

$$\tan\delta = \frac{B_v}{B_h},$$

where $$B_v$$ is the vertical component and $$\delta$$ is the angle that the total field makes with the horizontal.

Here the needle itself makes $$45^{\circ}$$ with the horizontal, so $$\delta = 45^{\circ}$$. Hence

$$\tan 45^{\circ} = 1 = \frac{B_v}{B_h} \quad\Longrightarrow\quad B_v = B_h = 18 \times 10^{-6}\ \text{T}.$$

Each pole of the magnet experiences a vertical magnetic force due to $$B_v$$ given by the basic formula

$$F_{\text{mag}} = m\,B_v.$$

For the north pole this force is, say, downward; for the south pole it is upward, so the two equal and opposite forces form a couple. The line joining the poles is the needle of length $$L$$, and the pivot is at the centre. The torque produced by this couple about the mid-point is obtained from the definition of torque of a couple (force × distance between the forces):

$$\tau_{\text{mag}} = m\,B_v \times L.$$

We now want the needle to stay exactly horizontal. For that we have to apply an external vertical force $$F$$ at one end of the needle. This force, acting at a perpendicular distance $$\dfrac{L}{2}$$ from the mid-point, produces an opposite torque

$$\tau_{\text{ext}} = F \times \frac{L}{2}.$$

In equilibrium the net torque must vanish, so

$$\tau_{\text{ext}} = \tau_{\text{mag}}.$$

Substituting the expressions, we get

$$F \times \frac{L}{2} = m\,B_v \times L.$$

Solving this for $$F$$ gives

$$F = \frac{m\,B_v \times L}{\dfrac{L}{2}} = 2\,m\,B_v.$$

Now we substitute the numerical values:

$$F = 2 \times 1.8\ \text{A m} \times 18 \times 10^{-6}\ \text{T}.$$

First multiply the ordinary numbers:

$$1.8 \times 18 = 32.4,$$

and then include the factor of 2:

$$2 \times 32.4 = 64.8.$$

Hence

$$F = 64.8 \times 10^{-6}\ \text{N} = 6.48 \times 10^{-5}\ \text{N}.$$

Rounding to two significant figures gives

$$F \approx 6.5 \times 10^{-5}\ \text{N}.$$

Hence, the correct answer is Option C.

The magnetic field at the midpoint P between the two dipoles is calculated using the formulas for axial and equatorial fields.

Let the dipole moment of X be M. Then the dipole moment of Y is 2M. The distance from each dipole to the midpoint P is r = d/2.

1. Magnetic field due to Dipole X (Axial point):

$$B_1 = \frac{\mu_0}{4\pi} \frac{2M}{(d/2)^3}$$

This field acts horizontally (along the axis of X).

2. Magnetic field due to Dipole $Y$ (Equatorial point):

$$B_2 = \frac{\mu_0}{4\pi} \frac{(2M)}{(d/2)^3}$$

This field acts vertically (perpendicular to the axis of Y).

3. Net Magnetic Field ($$B_{\text{net}}$$):

Since $$B_1 = B_2$$ and they are perpendicular to each other, the net magnetic field vector makes an angle with the horizontal:

$$\tan \phi = \frac{B_2}{B_1} = 1 \implies \phi = 45^\circ$$

4. Magnetic Force on the Particle:

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

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

$$F = qvB \sin \alpha$$

The particle is moving at an angle $$\theta = 45^\circ$$ with the horizontal, which is the same direction as $$\vec{B}_{\text{net}}$$. Therefore, the angle $$\alpha$$ between velocity and the magnetic field is $$0^\circ$$.

$$\sin(0^\circ) = 0$$

$$\boxed{F = 0}$$

Let us denote the magnetic dipole moment vector by $$\vec p$$ and the uniform magnetic field by $$\vec B$$. The angle between the two vectors is called $$\theta$$.

First, we recall two standard relations for a magnetic dipole kept in a uniform magnetic field:

1. Magnitude of torque

We have the vector formula $$\vec \tau = \vec p \times \vec B$$. Taking magnitudes, the cross-product gives

$$\tau = pB\sin\theta.$$

2. Potential energy

The scalar product of two vectors tells us

$$U = -\,\vec p \cdot \vec B.$$

This dot product becomes

$$U = -pB\cos\theta.$$

The question clearly states that the reference level of potential energy is chosen to be zero when the dipole is perpendicular to the field. Mathematically, when $$\theta = 90^{\circ}$$ we obtain

$$\cos 90^{\circ} = 0 \;\;\Longrightarrow\;\; U = -pB \times 0 = 0,$$

so the usual formula $$U = -pB\cos\theta$$ is already consistent with the given convention; no extra constant has to be added.

Now we compare the conditions for maximum torque and for the potential energy values:

Maximum torque

From $$\tau = pB\sin\theta$$ the sine term reaches its greatest value of 1 when

$$\sin\theta = 1 \;\;\Longrightarrow\;\; \theta = 90^{\circ}.$$

Hence the torque is maximum when the dipole is perpendicular to the field.

Potential energy at that same orientation

Substituting $$\theta = 90^{\circ}$$ in the energy expression, we get

$$U = -pB\cos 90^{\circ} = -pB \times 0 = 0.$$

So, exactly at the orientation where the torque attains its maximum value, the potential energy is zero (with the chosen reference point).

Among the given statements we therefore select the one that says “zero potential energy when the torque is maximum.”

Hence, the correct answer is Option B.

We know that a magnetic needle of magnetic moment $$M$$ placed in a uniform magnetic field $$B$$ executes torsional simple harmonic motion. The restoring torque on the needle is $$\tau = -MB\sin\theta$$. For very small angular displacements we can use the small-angle approximation $$\sin\theta \approx \theta$$, so the torque becomes $$\tau = -MB\,\theta$$.

Newton’s rotational equation is $$I\,\dfrac{d^{2}\theta}{dt^{2}} = \tau$$, where $$I$$ is the moment of inertia. Substituting the expression for $$\tau$$, we get

$$I\,\dfrac{d^{2}\theta}{dt^{2}} = -MB\,\theta.$$

This is the differential equation of a simple harmonic oscillator of the form $$\dfrac{d^{2}\theta}{dt^{2}} + \omega^{2}\theta = 0$$, where the square of the angular frequency is identified as

$$\omega^{2} = \dfrac{MB}{I}.$$

Therefore, the angular frequency is

$$\omega = \sqrt{\dfrac{MB}{I}}.$$

The time period $$T$$ of one complete oscillation is related to $$\omega$$ by the standard formula for SHM:

$$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{I}{MB}}.$$

Now we substitute the given numerical values. We have

$$M = 6.7 \times 10^{-2}\ \text{A m}^{2},\quad B = 0.01\ \text{T},\quad I = 7.5 \times 10^{-6}\ \text{kg m}^{2}.$$

First compute the product $$MB$$:

$$MB = \left(6.7 \times 10^{-2}\right)\left(0.01\right) = 6.7 \times 10^{-4}.$$

Next form the ratio $$\dfrac{I}{MB}$$:

$$\dfrac{I}{MB} = \dfrac{7.5 \times 10^{-6}}{6.7 \times 10^{-4}} = \dfrac{7.5}{6.7}\times 10^{-6+4} = 1.1194 \times 10^{-2}.$$

Taking the square root gives

$$\sqrt{\dfrac{I}{MB}} = \sqrt{1.1194 \times 10^{-2}} \approx 0.1058.$$

Now insert this value into the expression for the period:

$$T = 2\pi \left(0.1058\right) = 6.283 \times 0.1058 \approx 0.6647\ \text{s}.$$

This is the time for one complete oscillation. The question asks for the time taken for 10 complete oscillations, so we multiply by 10:

$$t_{10} = 10T = 10 \times 0.6647 \approx 6.647\ \text{s}.$$

Rounding to three significant figures, we obtain $$t_{10} \approx 6.65\ \text{s}.$$ Hence, the correct answer is Option B.

1. Material A:

Properties: This material has a large area within its hysteresis loop. It features high retentivity (residual magnetism when $$H=0$$) and high coercivity (resistance to being demagnetized).

Suitability: Because it is difficult to demagnetize and retains a strong magnetic field, it is ideal for making permanent magnets, such as those used in electric generators.

2. Material B:

Properties: This material has a narrow, thin loop. This indicates low coercivity and low hysteresis loss (the energy dissipated as heat during each cycle of magnetization).

Suitability: In devices like transformers and electromagnets, the magnetic field is constantly switching (alternating current). To prevent the device from overheating and to ensure high efficiency, you need a material that can be easily magnetized and demagnetized with minimal energy loss.

Applications: Specifically used for transformer cores and electromagnet cores.

Let the magnitudes of the two magnetic fields be denoted by $$B_1$$ and $$B_2$$. We are told

$$B_1 = 15\ \text{mT},\qquad B_2 = ?$$

The angle between the two fields is given as

$$\alpha = 75^\circ.$$

The magnetic dipole finally comes to rest (stable equilibrium) along the direction of the resultant field $$\mathbf R = \mathbf B_1 + \mathbf B_2$$. At equilibrium the dipole therefore makes the same angle with $$\mathbf B_1$$ as the resultant field does. This angle is stated to be

$$\theta = 30^\circ.$$

So, in the triangle formed by the vectors $$\mathbf B_1,\ \mathbf B_2,\ \mathbf R$$, we know the angle between $$\mathbf B_1$$ and $$\mathbf R$$ (namely $$\theta$$) and the angle between $$\mathbf B_1$$ and $$\mathbf B_2$$ (namely $$\alpha$$).

To relate the magnitudes we resolve $$\mathbf B_2$$ into components parallel and perpendicular to $$\mathbf B_1$$. Using elementary vector addition, the magnitude of the perpendicular component of $$\mathbf B_2$$ is

$$B_2 \sin\alpha,$$

while the component of $$\mathbf B_2$$ along $$\mathbf B_1$$ is

$$B_2 \cos\alpha.$$

The resultant $$\mathbf R$$ therefore has

• a component along $$\mathbf B_1$$ equal to $$B_1 + B_2\cos\alpha$$, and

• a component perpendicular to $$\mathbf B_1$$ equal to $$B_2\sin\alpha.$$

The angle $$\theta$$ that $$\mathbf R$$ makes with $$\mathbf B_1$$ is then obtained from simple trigonometry:

$$\tan\theta \;=\;\frac{\text{perpendicular component}}{\text{parallel component}} \;=\;\frac{B_2\sin\alpha}{B_1 + B_2\cos\alpha}.$$

We now substitute the known numerical values:

$$\tan 30^\circ \;=\;\frac{B_2\sin 75^\circ}{15 + B_2\cos 75^\circ}.$$

Next we insert the standard trigonometric numbers

$$\tan 30^\circ = \frac{1}{\sqrt 3} \approx 0.577,\qquad \sin 75^\circ \approx 0.9659,\qquad \cos 75^\circ \approx 0.2588.$$

This gives

$$0.577 =\frac{B_2(0.9659)}{15 + B_2(0.2588)}.$$

We now carry out the algebra step by step.

First cross-multiply:

$$0.577\bigl(15 + 0.2588\,B_2\bigr) = 0.9659\,B_2.$$

Multiply out the brackets on the left:

$$0.577 \times 15 + 0.577 \times 0.2588\,B_2 = 0.9659\,B_2.$$

Calculate the numerical products:

$$8.655 + 0.1493\,B_2 = 0.9659\,B_2.$$

Bring the $$B_2$$ terms together on one side:

$$8.655 = 0.9659\,B_2 - 0.1493\,B_2.$$

Combine the coefficients of $$B_2$$ on the right:

$$8.655 = (0.9659 - 0.1493)\,B_2 = 0.8166\,B_2.$$

Finally, solve for $$B_2$$ by dividing both sides by $$0.8166$$:

$$B_2 = \frac{8.655}{0.8166} \approx 10.6\ \text{mT}.$$

The closest option to $$10.6\ \text{mT}$$ is $$11\ \text{mT}$$.

Hence, the correct answer is Option B.

We are told that the neutral points lie on the East-West line drawn through the mid-point of the bar magnet. Such points are situated on the equatorial (broadside) line of the magnet. For a short bar magnet the magnetic induction at a point on the equatorial line, at a distance $$r$$ from the centre of the magnet, is given by the standard formula

$$B_{\text{equatorial}}=\frac{\mu_0}{4\pi}\,\frac{M}{r^{3}}$$

where $$M$$ is the magnetic moment of the magnet. This field is directed opposite to the direction of the earth’s horizontal field $$B_H$$. At a neutral point the two fields cancel in magnitude, so we must have

$$B_H = \frac{\mu_0}{4\pi}\,\frac{M}{r^{3}}$$

Now we solve for $$M$$. Rearranging the above equality gives

$$M = B_H \, r^{3}\,\left(\frac{4\pi}{\mu_0}\right)$$

We substitute the numerical values. The distance of the neutral point from the magnet’s centre is given as 30 cm, which in SI units is

$$r = 30\ \text{cm} = 0.30\ \text{m}$$

Cubing this distance, step by step,

$$r^{3} = (0.30\ \text{m})^{3} = 0.30 \times 0.30 \times 0.30 = 0.027\ \text{m}^{3}$$

The horizontal component of the earth’s magnetic field is

$$B_H = 3.6 \times 10^{-5}\ \text{T}$$

The constant $$\frac{\mu_0}{4\pi}$$ is given as $$10^{-7}$$ in SI units, hence

$$\frac{4\pi}{\mu_0} = \frac{1}{\mu_0/4\pi} = \frac{1}{10^{-7}} = 10^{7}$$

Substituting everything into the expression for $$M$$, we get

$$M = \left(3.6 \times 10^{-5}\right)\,\left(0.027\right)\,\left(10^{7}\right)$$

We first multiply the two decimal numbers:

$$3.6 \times 0.027 = 0.0972$$

Next we combine the powers of ten:

$$10^{-5} \times 10^{7} = 10^{2} = 100$$

Finally we multiply these results:

$$M = 0.0972 \times 100 = 9.72$$

So, the magnetic moment is

$$M \approx 9.7\ \text{A m}^2$$

Among the given choices, this value corresponds to Option D (9.7 A m2).

Hence, the correct answer is Option D.

We are given a solenoid with length 25 cm, radius 2 cm, 500 turns, and current 15 A. We need to find the magnitude of the magnetization $$\vec{M}$$, which is defined as the magnetic moment per unit volume. The solenoid is equivalent to a magnet of the same size, so we will calculate the magnetic moment of the solenoid and then divide by its volume.

First, convert all units to SI units. Length $$L = 25$$ cm = $$25 \times 10^{-2}$$ m = 0.25 m. Radius $$r = 2$$ cm = $$2 \times 10^{-2}$$ m = 0.02 m. Current $$I = 15$$ A. Number of turns $$N = 500$$.

The magnetic moment ($$m$$) of a solenoid is given by the product of the number of turns, the current, and the area of one turn. The area $$A$$ of one turn is the cross-sectional area of the solenoid, which is circular: $$A = \pi r^2$$.

Calculate $$A$$:

$$ A = \pi \times (0.02)^2 = \pi \times 0.0004 = 4 \times 10^{-4} \pi \text{ m}^2 $$

Now, calculate the magnetic moment $$m$$:

$$ m = N \times I \times A = 500 \times 15 \times (4 \times 10^{-4} \pi) $$

First, multiply 500 and 15:

$$ 500 \times 15 = 7500 $$

Then multiply by $$4 \times 10^{-4} \pi$$:

$$ 7500 \times 4 \times 10^{-4} \pi = 7500 \times 4 \times 0.0001 \pi = 7500 \times 0.0004 \pi $$

$$ 7500 \times 0.0004 = 3 \quad \text{(since } 7500 \times 4 \times 10^{-4} = 7500 \times 4 \times 0.0001 = 30000 \times 0.0001 = 3\text{)} $$

So,

$$ m = 3\pi \text{ A} \cdot \text{m}^2 $$

The volume ($$V$$) of the solenoid is the product of its cross-sectional area and length:

$$ V = A \times L = (4 \times 10^{-4} \pi) \times 0.25 $$

First, multiply 0.25 and $$4 \times 10^{-4}$$:

$$ 0.25 \times 4 \times 10^{-4} = 1 \times 10^{-4} = 0.0001 $$

So,

$$ V = 0.0001 \pi = \pi \times 10^{-4} \text{ m}^3 $$

Magnetization $$\vec{M}$$ is the magnetic moment per unit volume:

$$ |\vec{M}| = \frac{m}{V} = \frac{3\pi}{\pi \times 10^{-4}} $$

Simplify by canceling $$\pi$$:

$$ |\vec{M}| = \frac{3}{10^{-4}} = 3 \times 10^{4} = 30000 \text{ A m}^{-1} $$

Comparing with the options, 30000 A m$$^{-1}$$ corresponds to option B. The problem states that the correct answer is option 2, which is B.

Hence, the correct answer is Option B.

A superconductor acts as a perfect diamagnet by expelling all internal magnetic flux via the Meissner Effect, resulting in an internal magnetic field ($$B_s$$​) of zero.

We are told that the magnet will be demagnetised when it experiences an opposing magnetic field intensity (also called magnetic field strength) equal to its coercivity. Mathematically, the required magnetic field intensity is

$$H_c = 3 \times 10^3\ \text{A m}^{-1}.$$

The magnet is placed inside a long solenoid. For an ideal long solenoid, the magnetic field intensity $$H$$ produced at any interior point is related to the current $$I$$ flowing through it by the well-known relation

$$H = n I,$$

where $$n$$ is the number of turns per unit length of the solenoid.

Now we calculate $$n$$. The data given are:

Number of turns $$N = 100$$,   length of the solenoid $$L = 10\ \text{cm}.$$ We first convert the length to metres so that all quantities are in SI units:

$$L = 10\ \text{cm} = 10 \times 10^{-2}\ \text{m} = 0.10\ \text{m}.$$

Hence, the turn density is

$$n = \frac{N}{L} = \frac{100}{0.10} = 1000\ \text{turns m}^{-1}.$$

To demagnetise the magnet, the solenoid must supply an equal magnetic field intensity, so we set

$$H = H_c.$$

Substituting $$H = nI$$ and the numerical values, we get

$$n I = H_c$$ $$\Longrightarrow I = \frac{H_c}{n} = \frac{3 \times 10^3\ \text{A m}^{-1}}{1000\ \text{m}^{-1}} = 3\ \text{A}.$$

So, the current that has to be passed through the solenoid is

$$I = 3\ \text{A}.$$

Hence, the correct answer is Option C.

The magnetic field $$B$$ produced by the first dipole at the center of the second dipole along its axis is $$B = \frac{\mu_0}{4\pi} \frac{2M_1}{x^3}$$

$$U = -\vec{M_2} \cdot \vec{B} = -M_2 B \cos(0^\circ) = -\frac{\mu_0}{4\pi} \frac{2M_1 M_2}{x^3}$$

$$F = -\frac{dU}{dx} = -\frac{d}{dx} \left( -\frac{\mu_0}{4\pi} \frac{2M_1 M_2}{x^3} \right)$$

$$F = \frac{\mu_0}{4\pi} (2M_1 M_2) \frac{d}{dx} (x^{-3})$$

$$F = \frac{\mu_0}{4\pi} (2M_1 M_2) (-3x^{-4})$$

$$|F| \propto x^{-4}$$

$$n = 4$$

The magnetic field ($$B_{\text{eq}}$$) on the equatorial line of a dipole is given by:

$$B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{M}{R^3}$$

$$M = \frac{B_{\text{eq}} \cdot R^3}{\frac{\mu_0}{4\pi}}$$

$$M = \frac{4 \times 10^{-5} \times (6.4 \times 10^6)^3}{10^{-7}}$$

$$M \approx 1.05 \times 10^{23}\text{ A m}^2$$

Magnet A is allowing all the field lines passing through it , therefore, it is ferromagnetic. 

Magnet B is repelling the field lines, therefore, it is diamagnet. 

Magnet C is attracting the field lines, therefore, it is paramagnet.

The magnetic field at a distance $$R$$ on the equatorial line of a short dipole is given by:

$$B_{eq} = \frac{\mu_0}{4\pi} \frac{M}{R^3}$$

$$B_{eq} = 10^{-7} \times \frac{8 \times 10^{22}}{(6.4 \times 10^6)^3}$$

$$B_{eq} \approx 0.0305 \times 10^{-3}\text{ T} = 3.05 \times 10^{-5}\text{ T}$$

$$B_{eq} = 3.05 \times 10^{-5} \times 10^4\text{ Gauss}$$ ($$1\text{ Tesla} = 10^4\text{ Gauss}$$)

$$B_{eq} \approx 0.305\text{ Gauss} \approx 0.32\text{ Gauss}$$

We have two identical short magnets of length $$l = 1\;\text{cm} = 0.01\;\text{m}$$, but for the magnetic induction at a distant point only the magnetic moment is needed. Their magnetic moments are

$$M_1 = 1.20\; \text{A m}^2,$$

$$M_2 = 1.00\; \text{A m}^2.$$

The magnets are kept parallel to one another so that the line joining their centres is the common magnetic equator (broad-side line) for both of them. The distance between their centres is

$$d = 20.0\;\text{cm} = 0.20\;\text{m}.$$

Hence the mid-point $$O$$ of this line is at a distance

$$r = \dfrac{d}{2} = \dfrac{0.20\;\text{m}}{2} = 0.10\;\text{m}$$

from the centre of each magnet.

For a short bar magnet the magnetic induction at a point on its equatorial line is given by the standard formula

$$B_{\text{equator}} = \dfrac{\mu_0}{4\pi}\,\dfrac{M}{r^{3}},$$

where $$\mu_0/4\pi = 10^{-7}\;\text{T m A}^{-1}$$. The field is directed opposite to the magnetic moment vector. Since both magnets have their moments parallel, the fields produced by them at $$O$$ are along the same horizontal direction and hence add algebraically.

Magnetic induction at $$O$$ due to magnet 1:

$$B_1 = 10^{-7}\;\dfrac{M_1}{r^{3}} = 10^{-7}\;\dfrac{1.20}{(0.10)^3} = 10^{-7}\;\dfrac{1.20}{0.001} = 10^{-7}\times 1.20 \times 10^{3} = 1.20 \times 10^{-4}\;\text{T}.$$

Magnetic induction at $$O$$ due to magnet 2:

$$B_2 = 10^{-7}\;\dfrac{M_2}{r^{3}} = 10^{-7}\;\dfrac{1.00}{0.001} = 1.00 \times 10^{-4}\;\text{T}.$$

Adding these two gives the total field at $$O$$ because of the two magnets:

$$B_{\text{magnets}} = B_1 + B_2 = (1.20 + 1.00) \times 10^{-4} = 2.20 \times 10^{-4}\;\text{T}.$$

The horizontal component of the Earth’s magnetic induction is supplied as

$$B_{\text{Earth}} = 3.6 \times 10^{-5}\;\text{T} = 0.36 \times 10^{-4}\;\text{T}.$$

The equatorial fields of the magnets are directed towards geographic north (opposite to their moments), which is the same direction as the Earth’s horizontal field. Therefore the two contributions add directly. The resultant horizontal magnetic induction at $$O$$ is

$$B_{\text{resultant}} = B_{\text{magnets}} + B_{\text{Earth}} = 2.20 \times 10^{-4}\; \text{T} + 0.36 \times 10^{-4}\; \text{T} = 2.56 \times 10^{-4}\;\text{T}.$$

Expressing in the requested units of $$\text{Wb m}^{-2}$$ (which are identical to tesla), we have

$$B_{\text{resultant}} = 2.56 \times 10^{-4}\;\text{Wb m}^{-2}.$$

Hence, the correct answer is Option 4.