JEE Physics Magnetism and Matter MCQs Set 32

Answer: (a) \( \frac{\mu_0}{4\pi}\frac{M}{d^3} \)

Explanation:
1. At one of the other corners, one magnet is on its axis at distance d: \( \frac{\mu_0}{4\pi}\frac{2M}{d^3} \) along the moment.
2. The other lies on its equatorial line at distance d: \( \frac{\mu_0}{4\pi}\frac{M}{d^3} \) opposite to the moment.
3. Net \( = \frac{\mu_0}{4\pi}\frac{M}{d^3} \).

Teacher's Note:
1. Axial field is twice the equatorial field at the same distance.
2. The axial and equatorial fields are in opposite directions.

Answer: (d) 50 μJ

Explanation:
1. Taking θ as the angle between the moment and the horizontal field, \( U = -MB_H\cos\theta \). East-west position: θ = 90°, U = 0. North-south position: θ = 0, U = −MB_H.
2. Kinetic energy gained \( = MB_H = 2\times25\times10^{-6} = 50\ \mu\text{J} \).

Teacher's Note:
1. Torque does work as the magnet rotates through 90°.
2. Potential energy changes from zero to −MB.

Answer: (b) 45° and 10° W

Explanation:
1. \( \tan\delta = \frac{B_V}{B_H} = 1 \Rightarrow \delta = 45^\circ \).
2. The horizontal component points 10° west of geographic north, so the declination is 10° W.

Teacher's Note:
1. Declination is the angle between magnetic and geographic meridians.
2. Dip is given by tan δ = B_V/B_H.

Answer: (d) 596

Explanation:
1. \( B = \frac{2.4\times10^{-5}}{0.2\times10^{-4}} = 1.2 \) T.
2. \( \mu_r = \frac{B}{\mu_0H} = \frac{1.2}{4\pi\times10^{-7}\times1600} \approx 597 \); \( \chi = \mu_r-1 \approx 596 \).

Teacher's Note:
1. Convert cm² to m² before finding B.
2. Susceptibility χ = μ_r − 1.

Answer: (a) both a superconductor and a soft iron ring

Explanation:
1. A soft iron ring has very high permeability, so field lines are drawn into the iron and the space inside is shielded.
2. A superconductor expels the magnetic field (Meissner effect), so it also shields.

Teacher's Note:
1. Both diamagnetic expulsion and high permeability can shield.
2. Electrostatic shielding needs a conductor; magnetic shielding needs these.

Answer: (c) \( \tan^{-1}\left(\frac12\right) \)

Explanation:
1. Apparent dip: \( \tan\delta' = \frac{\tan\delta}{\cos\alpha} \).
2. \( \tan\delta = \tan30^\circ\cos30^\circ = \frac1{\sqrt3}\cdot\frac{\sqrt3}2 = \frac12 \).

Teacher's Note:
1. The apparent horizontal component is B_H cos α.
2. True dip is smaller than the apparent dip.

Answer: (c) \( \tan^{-1}\left(\frac12\right) \)

Explanation:
1. \( \cot^2\delta = \cot^2\delta_1+\cot^2\delta_2 = 3+1 = 4 \).
2. So \( \cot\delta = 2 \) and \( \delta = \tan^{-1}\frac12 \).

Teacher's Note:
1. For perpendicular meridians, cos²α + sin²α = 1 gives the cotangent-square relation.
2. Compute cot 30° = √3 and cot 45° = 1.