JEE Physics Magnetism and Matter MCQs Set 30

Answer: (a) energy is supplied to the electron

Explanation:
1. The magnetic moment of an electron is opposite to its angular momentum (\( \vec\mu=-\frac{e}{2m}\vec L \)).
2. With \( \vec L \) antiparallel to \( \vec B \), \( \vec\mu \) is parallel to \( \vec B \), the lowest energy state \( U=-\vec\mu\cdot\vec B \). After the flip \( \vec\mu \) is antiparallel to \( \vec B \) and \( U \) is higher, so energy must be supplied.

Teacher's Note:
1. A negative charge has its magnetic moment opposite to its angular momentum.

Answer: (a) 1 s

Explanation:
1. \( T\propto\frac1{\sqrt M} \) (same moment of inertia).
2. \( M'=4M \) gives \( T'=\frac T2=1 \) s.

Teacher's Note:
1. The time period varies inversely as the square root of the magnetic moment.

Answer: (b) a ferromagnetic material becomes paramagnetic

Explanation:
1. Above the Curie temperature thermal agitation destroys the domains, so a ferromagnetic material behaves as a paramagnetic one.

Teacher's Note:
1. Domain alignment is lost above the Curie temperature.

Answer: (d) behaves like a paramagnetic material

Explanation:
1. The domain structure is destroyed above the Curie temperature and the material becomes paramagnetic.

Teacher's Note:
1. Ferromagnetism disappears above the Curie point.

Answer: (d) \( \frac{\mu_0}{4\pi}\sqrt5\frac{M}{d^3} \)

Explanation:
1. The mid-point is at distance \( d \) from each. It is on the axial line of one magnet: \( B_1=\frac{\mu_0}{4\pi}\frac{2M}{d^3} \), and on the equatorial line of the other: \( B_2=\frac{\mu_0}{4\pi}\frac{M}{d^3} \).
2. The fields are perpendicular: \( B=\frac{\mu_0}{4\pi}\frac{M}{d^3}\sqrt{4+1}=\frac{\mu_0}{4\pi}\sqrt5\frac M{d^3} \).

Teacher's Note:
1. Add the axial and equatorial fields as perpendicular vectors.

Answer: (b) \( \tan^{-1}2 \)

Explanation:
1. For a plane at angle \( \alpha \) to the meridian, \( \tan\delta'=\frac{\tan\delta}{\cos\alpha} \).
2. \( \tan\delta'=\frac{\tan60^\circ}{\cos30^\circ}=\frac{\sqrt3}{\sqrt3/2}=2 \).

Teacher's Note:
1. In an inclined plane only B_H cos α acts horizontally, while B_V is unchanged.

Answer: (b) (ii) and (iv) only

Explanation:
1. A diamagnetic body acquires a dipole moment opposite to the field. Near the south pole the field points towards the magnet, so the induced moments point away from it: (iv) true, (iii) false.
2. The induced dipole is repelled from the stronger field, so the force on each sphere is away from the magnet: (ii) true, (i) false.

Teacher's Note:
1. Diamagnetic bodies are repelled by both poles of a magnet.

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

Explanation:
1. Diamagnetism is a universal weak effect caused by induced moments opposing the applied field: (i) and (iv) are true.
2. It does not need permanent moments, so (ii) is false; and \( H=\frac{B}{\mu_0} \) in free space is not zero in a field, so (iii) is false.

Teacher's Note:
1. Induced moments in diamagnetic materials always oppose the field.

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

Explanation:
1. Curie's law: \( \chi_m=\frac{C}{T} \), which is positive and decreases as \( T \) rises.

Teacher's Note:
1. Paramagnetism follows Curie's law.

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

Explanation:
1. Magnetic field lines form closed loops, so as many lines enter a closed surface as leave it: net flux is zero.
2. This is a consequence of the absence of isolated poles, so R explains A.

Teacher's Note:
1. The magnetic counterpart of Gauss's law says ∮B·dA = 0.

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

Explanation:
1. Poles always occur in pairs, so A is true.
2. R is also true, but it is not the reason for A.

Teacher's Note:
1. Magnetic poles occur in north-south pairs.

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

Explanation:
1. \( \tan\theta=\frac{\mu_0Ni}{2rB_H} \), so for a given current the deflection grows with \( N \). A is true.
2. For a given deflection, \( i\propto\frac1N \), so R is false.

Teacher's Note:
1. Sensitivity grows with N; the current needed for a given deflection falls with N.

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

Explanation:
1. A soft iron core concentrates the field and makes the radial field stronger, so A is true.
2. R is false: soft iron has high permeability and can be easily magnetised and demagnetised.

Teacher's Note:
1. Soft iron magnetises and demagnetises very easily.

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

Explanation:
1. \( \tan\delta=\frac{B_V}{B_H}=\frac{1}{\sqrt3} \), so \( \delta=30^\circ \).

Teacher's Note:
1. Dip is given by the ratio of the vertical to the horizontal component.

Answer: (a) \( 50\ \mu \)J

Explanation:
1. Work done by the field \( =MB_H(\cos0^\circ-\cos90^\circ)=2.0\times25\times10^{-6}=5\times10^{-5} \) J.
2. This equals the kinetic energy gained, \( 50\ \mu \)J.

Teacher's Note:
1. Only the horizontal component of the field acts on a magnet rotating in a horizontal plane.

Answer: (a) \( 25\ \text{A m}^2 \)

Explanation:
1. \( n=500 \) turns/m, so \( H=ni=250 \) A/m.
2. \( \chi=\mu_r-1=999 \), magnetisation \( I=\chi H\approx2.5\times10^5 \) A/m.
3. \( M=IV=2.5\times10^5\times10^{-4}\approx25\ \text{A m}^2 \).

Teacher's Note:
1. Magnetic moment is magnetisation times volume.

Answer: (d) \( 0.18\ \text{A/m} \)

Explanation:
1. \( n=4000 \) turns/m, so \( H=ni=8000 \) A/m.
2. \( I=\chi H=2.3\times10^{-5}\times8000=0.18 \) A/m.

Teacher's Note:
1. Magnetisation is susceptibility times the magnetising field.

Answer: (a) \( 1.6\times10^{-7} \) T

Explanation:
1. \( B=\frac{\mu_0}{4\pi}\frac{M}{r^3}\sqrt{1+3\cos^2\theta}=10^{-7}\times1.2\times\sqrt{1+0.75} \).
2. \( =1.2\times10^{-7}\times1.32=1.6\times10^{-7} \) T.

Teacher's Note:
1. Use B = (μ0 M / 4π r³) √(1 + 3cos²θ).

Answer: (d) \( \tan^{-1}\left(\frac{\sqrt3}{8}\right) \)

Explanation:
1. On the axial line \( B\propto\frac1{d^3} \) and \( \tan\theta=\frac{B}{B_H} \).
2. Doubling \( d \) divides \( \tan\theta \) by 8: \( \tan\theta'=\frac{\tan60^\circ}{8}=\frac{\sqrt3}{8} \).

Teacher's Note:
1. In the tan-A position the deflection obeys tan θ ∝ 1/d³.

Answer: (b) \( 6.9\times10^{-3}\ \text{T m/A} \)

Explanation:
1. \( \mu=\mu_0(1+\chi)=4\pi\times10^{-7}\times5501=6.9\times10^{-3} \) T m/A.

Teacher's Note:
1. Permeability is μ0 times (1 + χ).