JEE Physics Magnetism and Matter MCQs Set 29

Answer: (c) will stay in any position

Explanation:
1. At a geomagnetic pole the earth's field is vertical (dip \( =90^\circ \)), so the horizontal component \( B_H=B\cos\delta=0 \).
2. With no horizontal field there is no torque on a needle confined to a horizontal plane, so it stays in any position.

Teacher's Note:
1. A compass works only because of the horizontal component of the earth's field.

Answer: (b) \( \frac{2M}{\pi} \)

Explanation:
1. \( M=m\times L \) for pole strength \( m \). After bending, the poles are at the ends of the semicircle, a distance equal to its diameter \( 2R \) apart, with \( \pi R=L \).
2. \( M'=m\cdot2R=m\frac{2L}{\pi}=\frac{2M}{\pi} \).

Teacher's Note:
1. Magnetic moment is pole strength times the straight-line distance between the poles.

Answer: (d) zero

Explanation:
1. Work done is the change in potential energy, \( U=-MH\cos\theta \).
2. After a full rotation the magnet returns to its original orientation, so the change in \( U \) is zero and the work done is zero.

Teacher's Note:
1. The magnetic potential energy depends only on the final and initial orientation.

Answer: (b) \( 150\hat k \) N m

Explanation:
1. \( \vec\tau=\vec M\times\vec B=50\hat i\times(0.5\hat i+3.0\hat j) \).
2. \( \hat i\times\hat i=0 \) and \( \hat i\times\hat j=\hat k \), so \( \vec\tau=150\hat k \) N m.

Teacher's Note:
1. Only the component of B perpendicular to M produces a torque.

Answer: (c) \( 45^\circ \)

Explanation:
1. \( \tan\delta=\frac{B_V}{B_H}=1 \), so \( \delta=45^\circ \).

Teacher's Note:
1. The tangent of the dip angle is the ratio of the vertical to the horizontal component.

Answer: (c) \( \sqrt2\,T_0 \)

Explanation:
1. The size and pole strength are unchanged, so \( M \) is unchanged. Double the mass doubles the moment of inertia.
2. \( T\propto\sqrt I \), so \( T'=\sqrt2\,T_0 \).

Teacher's Note:
1. Only the moment of inertia changes here.

Answer: (b) 5499

Explanation:
1. \( \chi=\mu_r-1=5500-1=5499 \).

Teacher's Note:
1. Susceptibility is one less than the relative permeability.

Answer: (b) paramagnetic

Explanation:
1. Liquid oxygen is attracted into the strong field region, so it is paramagnetic.

Teacher's Note:
1. Paramagnetic substances move towards regions of stronger field.

Answer: (a) \( \tan^{-1}\left(\frac{H_0}{H}\right) \)

Explanation:
1. The coil's field \( H \) is along its axis (east-west) and \( H_0 \) is north-south, so they are perpendicular.
2. The magnet lines up with the resultant, which makes an angle \( \theta \) with \( H \) where \( \tan\theta=\frac{H_0}{H} \).

Teacher's Note:
1. Two perpendicular fields: the angle with one is the inverse tangent of the other over it.

Answer: (d) all the above

Explanation:
1. Jerks disturb the domain alignment, heating destroys it, and a reverse field realigns the domains the other way. All three demagnetise a magnet.

Teacher's Note:
1. Anything that disorders the aligned dipoles removes the magnetisation.

Answer: (b) \( \frac23 \) s

Explanation:
1. Each piece keeps the pole strength but has one third of the length, so its moment is \( \frac M3 \); the combination has moment \( 3\times\frac M3=M \).
2. The moment of inertia of each piece (a rod of mass \( \frac m3 \), length \( \frac l3 \)) is \( \frac{1}{3}\cdot\frac1{9}I=\frac{I}{27} \). Three pieces side by side (width neglected) give \( I'=\frac I9 \).
3. \( T\propto\sqrt{\frac IM} \), so \( T'=\frac T3=\frac23 \) s.

Teacher's Note:
1. Find the new moment of inertia and the new magnetic moment separately.

Answer: (a) 0.4 T

Explanation:
1. The coil is a dipole of moment \( NiA \) perpendicular to the field, so the torque is \( NiAB \).
2. Torque balance: \( NiAB=\Delta m\,g\,\ell \), so \( B=\frac{60\times10^{-6}\times9.8\times0.3}{200\times22\times10^{-3}\times10^{-4}}=\frac{1.764\times10^{-4}}{4.4\times10^{-4}}=0.4 \) T.

Teacher's Note:
1. Equate the magnetic torque on the coil with the torque of the counter-weight.

Answer: (c) \( \frac{1}{\sqrt{(B_1\cos\alpha)^2+B_2^2}} \)

Explanation:
1. In the vertical plane of oscillation the field has the horizontal component \( B_1\cos\alpha \) and the vertical component \( B_2 \).
2. The effective field in that plane is \( \sqrt{(B_1\cos\alpha)^2+B_2^2} \), so \( T^2\propto\frac{1}{B_{eff}}=\frac{1}{\sqrt{(B_1\cos\alpha)^2+B_2^2}} \).

Teacher's Note:
1. A needle oscillating in a plane feels the component of the earth's field lying in that plane.

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

Explanation:
1. A is false: the core is far above the Curie temperature of iron, so it cannot be a permanent magnet. The field is attributed to currents in the molten outer core.
2. R is true (above the Curie temperature ferromagnetism disappears).

Teacher's Note:
1. The earth's core is too hot to be a permanent magnet.

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

Explanation:
1. A permanent magnet must retain its magnetism, which requires high retentivity (and high coercivity). Steel has both, so R explains A.

Teacher's Note:
1. Permanent magnets need materials with wide hysteresis loops.

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

Explanation:
1. A transformer core is magnetised and demagnetised many times a second, so low hysteresis loss is required. Soft iron has a narrow loop and satisfies this, so R explains A.

Teacher's Note:
1. Hysteresis loss is proportional to the area of the B-H loop.

Answer: (d) \( \tan\delta''=\sqrt{\frac32} \)

Explanation:
1. \( \tan60^\circ=\frac{\tan45^\circ}{\cos\alpha} \) gives \( \cos\alpha=\frac1{\sqrt3} \), \( \sin\alpha=\sqrt{\frac23} \).
2. After rotating through \( 90^\circ \) the plane makes \( 90^\circ-\alpha \) with the meridian: \( \tan\delta''=\frac{1}{\cos(90^\circ-\alpha)}=\frac{1}{\sin\alpha}=\sqrt{\frac32} \) (so \( \cot\delta''\approx0.816 \)).

Teacher's Note:
1. Rotating the dip circle by 90° swaps cos α for sin α in the apparent-dip formula.

Answer: (d) \( 4.0\times10^{-5} \) N m/rad

Explanation:
1. Equilibrium: \( NiAB=C\theta \).
2. \( C=\frac{100\times10^{-2}\times2\times10^{-4}\times0.01}{0.05}=4\times10^{-5} \) N m/rad.

Teacher's Note:
1. Magnetic torque NiAB balances the restoring torque Cθ.

Answer: (b) \( 2.5\times10^{-2} \) N

Explanation:
1. For long magnets only the nearest poles matter: \( F=\frac{\mu_0}{4\pi}\frac{m_1m_2}{r^2}=10^{-7}\times\frac{10\times10}{(0.02)^2}=2.5\times10^{-2} \) N (attractive).

Teacher's Note:
1. Treat each long magnet as an isolated pole at the facing end.

Answer: (c) \( 1.58\times10^{6} \) A/m

Explanation:
1. Maximum magnetisation (all moments aligned): \( I=n\mu=8.52\times10^{28}\times2\times9.27\times10^{-24}=1.58\times10^{6} \) A/m.

Teacher's Note:
1. Saturation magnetisation is the number density times the moment per atom.