JEE Physics Magnetism and Matter MCQs Set 28

Answer: (b) M/2

Explanation:
1. Cutting across the length halves the length and keeps the pole strength.
2. \( M'=m\times\tfrac L2=\tfrac M2 \).

Teacher's Note:
1. Magnetic moment is pole strength times length.
2. Cutting across the magnet halves only the length.

Answer: (a) 38.4 A m\(^2\)

Explanation:
1. \( V=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^2} \).
2. \( M=\dfrac{Vr^2}{10^{-7}}=\dfrac{2.4\times10^{-5}\times0.16}{10^{-7}}=38.4 \) A m\(^2\).

Teacher's Note:
1. Use the dipole potential formula with mu-zero over 4 pi equal to 10 to the power minus 7 and r in metres.

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

Explanation:
1. \( W=MB(1-\cos60^\circ)=\tfrac12MB \), so \( MB=2W \).
2. \( \tau=MB\sin60^\circ=2W\times\tfrac{\sqrt3}2=\sqrt3\,W \).

Teacher's Note:
1. Find MB from the work done, then use the torque at the new angle.

Answer: (a) 8

Explanation:
1. For a short magnet on the axis, \( B\propto1/r^3 \).
2. \( B_A/B_B=(48/24)^3=8 \).

Teacher's Note:
1. The axial field of a short magnet falls off as the cube of the distance.

Answer: (a) 0°

Explanation:
1. \( \tau=MB\sin\theta \), so \( \dfrac{d\tau}{d\theta}=MB\cos\theta \).
2. This is maximum at \( \theta=0^\circ \).

Teacher's Note:
1. Differentiate the torque.
2. The cosine is largest at zero angle.

Answer: (b) 12 dyne

Explanation:
1. End-on force between short magnets: \( F=\dfrac{\mu_0}{4\pi}\dfrac{6M_1M_2}{r^4} \), with \( \mu_0/4\pi=1 \) in CGS.
2. \( F=\dfrac{6\times400\times800}{20^4}=\dfrac{1.92\times10^6}{1.6\times10^5}=12 \) dyne.

Teacher's Note:
1. In CGS units the factor mu-zero over 4 pi is one.
2. The force of two coaxial dipoles falls as the fourth power of distance.

Answer: (c) both have equal magnetic moment

Explanation:
1. Magnetic moment = pole strength times magnetic length.
2. Both quantities are the same, so the moments are equal.

Teacher's Note:
1. A small hole at the centre does not change the pole strength or length, so the moment is unchanged.

Answer: (d) 0.36 J/T

Explanation:
1. \( \tau=MB\sin\theta \).
2. \( M=\dfrac{4.5\times10^{-2}}{0.25\times0.5}=0.36 \) J/T.

Teacher's Note:
1. Rearrange the torque formula for the moment, using sin 30 degrees equal to one half.

Answer: (b) 0.6 J/T

Explanation:
1. A current-carrying solenoid acts like a bar magnet with moment \( M=NIA \).
2. \( M=800\times3.0\times2.5\times10^{-4}=0.6 \) J/T.

Teacher's Note:
1. Magnetic moment of a coil is the number of turns times current times area.

Answer: (a) intensity of magnetisation

Explanation:
1. Cutting along the length halves the pole strength and the magnetic moment, and halves the volume.
2. Intensity of magnetisation \( I=M/V \) is unchanged.

Teacher's Note:
1. Moment per unit volume is unchanged when the magnet is cut along its length.

Answer: (d) the magnetic equator

Explanation:
1. Angle of dip is 0° at the magnetic equator.
2. The vertical component \( B_V=B\sin\delta=0 \) there.

Teacher's Note:
1. At the magnetic equator the dip needle is horizontal.

Answer: (a) \( \tan^{-1}\dfrac43 \)

Explanation:
1. \( B_V=\sqrt{0.5^2-0.3^2}=0.4 \) G.
2. \( \tan\delta=B_V/B_H=0.4/0.3=4/3 \).

Teacher's Note:
1. Find the vertical component from the total and horizontal components, then take their ratio for the tangent of the dip angle.

Answer: (a) east and west of the magnet

Explanation:
1. With its N pole pointing north, the magnet's field at points on its equatorial line points from N to S, that is southward.
2. The earth's horizontal field points northward, so they cancel on the equatorial line, which lies east and west.

Teacher's Note:
1. Neutral points lie on the equatorial line when the magnet's north pole points north.

Answer: (d) 45 A m

Explanation:
1. With the south pole towards north, the neutral points lie on the axial line.
2. \( B_H=\dfrac{\mu_0}{4\pi}\dfrac{2Md}{(d^2-l^2)^2} \) with \( M=m(2l)=0.2m \), \( d=0.4 \), \( l=0.1 \).
3. \( 3.2\times10^{-5}=10^{-7}\dfrac{2(0.2m)(0.4)}{(0.15)^2}=7.11\times10^{-7}m \), so \( m=45 \) A m.

Teacher's Note:
1. South pole pointing north places the neutral points on the axis.
2. Use the axial field formula with half-length l.

Answer: (a) \( \sqrt2\,B_0 \)

Explanation:
1. \( B_H=B\cos\delta \).
2. \( B=B_0/\cos45^\circ=\sqrt2\,B_0 \).

Teacher's Note:
1. The horizontal component is the total field times cosine of the dip.

Answer: (a) \( 6.28\ \text{T} \)

Explanation:
1. Volume \( V=10\times0.5\times0.2\ \text{cm}^3=1\ \text{cm}^3=10^{-6} \) m\(^3\).
2. \( I=M/V=5\times10^6 \) A/m.
3. \( B=\mu_0(H+I)=4\pi\times10^{-7}(5\times10^6+5\times10^3)\approx6.3 \) T.

Teacher's Note:
1. Intensity of magnetisation is moment per volume, and the induction is mu-zero times the sum of H and I.

Answer: (c) 0.0068%

Explanation:
1. \( B=\mu_0(1+\chi)H \), so the fractional increase is \( \chi \).
2. \( \chi\times100=6.8\times10^{-3}\% =0.0068\% \).

Teacher's Note:
1. The relative increase in field equals the susceptibility.

Answer: (b) 13.5 A/m

Explanation:
1. \( H=\dfrac{B}{\mu_0}=\dfrac{1.7\times10^{-5}}{4\pi\times10^{-7}} \).
2. \( H\approx13.5 \) A/m.

Teacher's Note:
1. Magnetic intensity in air is flux density divided by mu-zero.

Answer: (a) T/4

Explanation:
1. Each part has \( M'=M/4 \) and moment of inertia \( I'=\dfrac{(m/4)(l/4)^2}{12}=\dfrac{I}{64} \).
2. \( T\propto\sqrt{I/M} \), so \( T'=T\sqrt{\dfrac{1/64}{1/4}}=\dfrac T4 \).

Teacher's Note:
1. Both the moment of inertia and the magnetic moment change.
2. The moment of inertia falls by a factor of 64.

Answer: (a) \( 9\times10^{-6}\ \text{T} \)

Explanation:
1. \( f\propto\sqrt{B_H} \), so \( B_H\propto f^2 \).
2. \( B_B=36\times10^{-6}\times(20/40)^2=9\times10^{-6} \) T.

Teacher's Note:
1. Frequency varies as the square root of the field, so the field varies as the square of the frequency.