JEE Physics Magnetism and Matter MCQs Set 31

Answer: (a) a diamagnetic substance

Explanation:
1. Diamagnetic materials are magnetised opposite to the applied field, so \( \chi \lt 0 \).
2. Paramagnetic and ferromagnetic materials have \( \chi \gt 0 \).

Teacher's Note:
1. The sign of susceptibility tells whether the induced magnetisation is along or against the field.

Answer: (a) \( \sqrt3\,W \)

Explanation:
1. \( W=MB(1-\cos60^\circ)=\frac{MB}{2} \), so \( MB=2W \).
2. Torque at \( 60^\circ \): \( MB\sin60^\circ=2W\cdot\frac{\sqrt3}{2}=\sqrt3\,W \).

Teacher's Note:
1. First find MB from the work done, then use τ = MB sin θ.

Answer: (d) 0.3 N

Explanation:
1. For short magnets on a common axis the force varies as \( \frac1{r^4} \).
2. Doubling \( r \) reduces the force by \( 2^4=16 \): \( \frac{4.8}{16}=0.3 \) N.

Teacher's Note:
1. Dipole-dipole force along the axis falls off as the inverse fourth power.

Answer: (a) \( 4:9 \)

Explanation:
1. Frequency \( n\propto\sqrt{M} \) for equal moments of inertia.
2. \( \frac{M_1}{M_2}=\left(\frac{n_1}{n_2}\right)^2=\left(\frac{10}{15}\right)^2=\frac49 \).

Teacher's Note:
1. Equal size and mass means equal moment of inertia, so only M differs.

Answer: (c) the time period increases

Explanation:
1. Heating weakens the magnetism of the needle, so \( M \) decreases.
2. \( T=2\pi\sqrt{\frac{I}{MB_H}} \) increases when \( M \) decreases. (The needle does not lose its magnetism completely at this temperature.)

Teacher's Note:
1. Raising the temperature reduces the magnetic moment.

Answer: (b) \( 0.36\times10^{-6} \) T

Explanation:
1. At the first place \( T_1=\frac{60}{40}=1.5 \) s. \( T\propto\frac1{\sqrt{B_H}} \), so \( B_2=B_1\left(\frac{T_1}{T_2}\right)^2 \).
2. \( B_2=0.1\times10^{-5}\times\left(\frac{1.5}{2.5}\right)^2=0.36\times10^{-6} \) T.

Teacher's Note:
1. Convert oscillations per minute to a period before comparing.

Answer: (d) 0.1 m

Explanation:
1. Force on a pole \( F=mB \), so \( m=\frac{6\times10^{-4}}{2\times10^{-5}}=30 \) A m.
2. \( M=m\,l \) gives \( l=\frac{3.0}{30}=0.1 \) m.

Teacher's Note:
1. Find the pole strength from the force, then use M = m × length.

Answer: (b) \( \frac{eJ}{2m} \)

Explanation:
1. \( \mu=iA=\frac{e}{T}\pi r^2=\frac{e}{2}\,vr \) and \( J=mvr \).
2. So \( \mu=\frac{eJ}{2m} \).

Teacher's Note:
1. The ratio of magnetic moment to angular momentum is e/2m.

Answer: (d) \( \frac{\mu_0}{4\pi}\frac{2\sqrt2M}{d^3} \)

Explanation:
1. Take the moments along \( -\hat x \) and \( +\hat y \); the point lies along \( \hat r=\frac{(-1,1)}{\sqrt2} \) at distance \( d \). For one dipole \( \vec B=\frac{\mu_0}{4\pi d^3}\left[3(\vec M\cdot\hat r)\hat r-\vec M\right] \).
2. With \( \vec M\cdot\hat r=\frac{M}{\sqrt2} \) for each: \( \vec B_1\propto M(-0.5,1.5) \) and \( \vec B_2\propto M(-1.5,0.5) \).
3. The sum is \( M(-2,2) \), of magnitude \( 2\sqrt2M \), so \( B=\frac{\mu_0}{4\pi}\frac{2\sqrt2M}{d^3} \).

Teacher's Note:
1. Use the general dipole-field formula with vectors when the point is neither axial nor equatorial.

Answer: (a) nearly \( 2\pi aiB \) perpendicular to the plane of the loop

Explanation:
1. The magnet's field at the loop is radial (in the plane of the loop) because the pole is at the centre. The current element \( i\,d\vec l \) is tangential, so \( d\vec F=i\,d\vec l\times\vec B \) is perpendicular to the plane of the loop.
2. All elements give force in the same direction, so \( F=i(2\pi a)B \) perpendicular to the plane.

Teacher's Note:
1. The force on a current element is perpendicular to both the element and the field.

Answer: (a) (i) and (ii) only

Explanation:
1. Domains that are aligned with the field grow at the expense of misaligned ones, so some domains increase and some decrease in size. Their size is not constant.

Teacher's Note:
1. The field makes favourably oriented domains grow and unfavourable ones shrink.

Answer: (a) (i), (ii) and (iii)

Explanation:
1. Current sensitivity \( \frac{\theta}{I}=\frac{NAB}{C} \). It increases with \( N \), \( A \) and \( B \) and decreases when the torsional constant \( C \) (the restoring couple) increases.

Teacher's Note:
1. Current sensitivity is NAB divided by the torsional constant.

Answer: (c) A is true, but R is false.

Explanation:
1. \( i=K\tan\theta \) with \( K=\frac{2rB_H}{\mu_0N} \), so A is true.
2. \( K \) is a constant of the instrument and does not depend on the current, so R is false.

Teacher's Note:
1. The reduction factor depends on r, N and B_H only.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. All atoms have electrons whose orbits respond to a field, giving a diamagnetic effect; in paramagnetic substances this is masked by the much larger paramagnetic effect. Both statements are true and R explains A.

Teacher's Note:
1. Diamagnetism is present in every material.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: a ferromagnet follows the Curie-Weiss law above \( T_C \) and not \( \chi=\frac CT \).
2. R is true but does not by itself explain why Curie's law is not obeyed.

Teacher's Note:
1. Ferromagnets follow the Curie-Weiss law above the Curie temperature.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The field lines prefer the high-permeability iron path and are diverted around the enclosed space, so R explains A.

Teacher's Note:
1. Magnetic shielding works by diverting field lines through high-permeability material.

Answer: (c) \( 30^\circ \)

Explanation:
1. \( \tan\delta'=\frac{\tan\delta}{\cos\alpha} \): \( \frac{2}{\sqrt3}=\frac{\tan\delta}{\cos60^\circ} \), so \( \tan\delta=\frac1{\sqrt3} \).
2. \( \delta=30^\circ \).

Teacher's Note:
1. Apparent dip in an inclined vertical plane is tan δ divided by cos α.

Answer: (a) 570

Explanation:
1. At \( 45^\circ \), \( \tan\theta=1 \): \( B_H=\frac{\mu_0Ni}{2r} \).
2. \( N=\frac{2rB_H}{\mu_0i}=\frac{2\times0.1\times3.6\times10^{-5}}{4\pi\times10^{-7}\times10^{-2}}\approx573\approx570 \).

Teacher's Note:
1. At 45° the coil's field equals the horizontal component of the earth's field.

Answer: (d) \( \frac{16}{9\sqrt3} \)

Explanation:
1. The frequency satisfies \( n^2\propto B_H=B\cos\delta \).
2. \( \frac{B_1}{B_2}=\frac{n_1^2\cos\delta_2}{n_2^2\cos\delta_1}=\frac{400\cos60^\circ}{225\cos30^\circ}=\frac{400\times0.5}{225\times0.866}=\frac{16}{9\sqrt3} \).

Teacher's Note:
1. Compare B cos δ from the squares of the frequencies.

Answer: (b) \( 3.4\times10^{-5} \) T

Explanation:
1. The poles are at distances \( 0.15 \) m and \( 0.25 \) m from the point (half-length \( 0.05 \) m).
2. \( B=\frac{\mu_0}{4\pi}m\left[\frac1{0.15^2}-\frac1{0.25^2}\right]=10^{-7}\times12\times(44.4-16)=3.4\times10^{-5} \) T.

Teacher's Note:
1. Add the fields of the two poles separately when the magnet is not short.