JEE Physics Magnetism and Matter MCQs Set 37

Answer: (c) diamagnetic materials

Explanation:
1. Diamagnetic materials are weakly repelled by a magnetic field and acquire magnetisation opposite to the field.
2. So their susceptibility \( \chi_m \) is small and negative.
3. Paramagnetic and ferromagnetic materials have positive susceptibility.

Teacher's Note:
1. Negative susceptibility means magnetisation opposes the field.
2. Remember: dia is negative, para is small positive, ferro is large positive.

Answer: (c) \( H \)

Explanation:
1. Magnetic susceptibility is defined by \( \chi_m = \frac{I}{H} \).
2. For a small field \( \chi_m \) is constant for the material.
3. Hence \( I \propto H \).

Teacher's Note:
1. Use the definition \( I = \chi_m H \).
2. The proportionality holds for weak fields, before saturation.

Answer: (b) \( \chi \propto T^{-1} \)

Explanation:
1. For a paramagnetic substance, Curie's law states \( \chi = \frac{C}{T} \).
2. Increasing temperature randomises the atomic dipoles and lowers the magnetisation.
3. So \( \chi \propto T^{-1} \).

Teacher's Note:
1. Curie's law is for paramagnets; Curie-Weiss law applies to ferromagnets above their Curie point.
2. Higher temperature means weaker alignment.

Answer: (b) (a) is true but (b) is false

Explanation:
1. Magnetisation of a paramagnet grows with the applied field (until saturation): (a) is true.
2. Thermal motion disturbs the alignment of the dipoles, so magnetisation decreases with rising temperature (\( I \propto \frac{B}{T} \)): (b) is false.
3. So only (a) is true.

Teacher's Note:
1. Use \( I \propto \frac{B}{T} \) for paramagnets.
2. Temperature works against alignment.

Answer: (c) \( \mu_0 = \mu(1 + \chi_m) \)

Explanation:
1. \( B = \mu_0(H + I) \) and, with \( I = \chi_m H \), \( B = \mu_0 H(1 + \chi_m) \): (a) and (b) are correct.
2. Since \( B = \mu H \), \( \mu = \mu_0(1 + \chi_m) \) and \( \mu_r = 1 + \chi_m \): (d) is correct.
3. (c) is wrong: the correct relation is \( \mu = \mu_0(1 + \chi_m) \), not \( \mu_0 = \mu(1 + \chi_m) \).

Teacher's Note:
1. Start from \( B = \mu_0(H + I) \) and \( I = \chi_m H \).
2. \( \mu \) is the permeability of the material and \( \mu_0 \) that of vacuum.

Answer: (c) tall and wide

Explanation:
1. A permanent magnet needs high retentivity (the loop extends far up the B axis, so it is tall).
2. It also needs high coercivity so that it is not easily demagnetised (the loop extends far along the H axis, so it is wide).
3. So the loop is tall and wide.

Teacher's Note:
1. Tall means large retentivity; wide means large coercivity.
2. Soft iron has the opposite loop: tall and narrow.

Answer: (a) It has a fixed value.

Explanation:
1. Along the hysteresis loop the ratio \( \frac{I}{H} \) (or the slope \( \frac{dI}{dH} \)) changes from point to point.
2. It can be zero (at saturation), infinite (where \( H = 0 \) but \( I \ne 0 \)) and negative (on the descending part of the curve).
3. So the claim that it has a fixed value is incorrect.

Teacher's Note:
1. A ferromagnet does not follow a single linear relation between \( I \) and \( H \).
2. Think of the shape of the hysteresis loop.

Answer: (a) high retentivity and high coercivity

Explanation:
1. A permanent magnet must keep strong magnetism after the field is removed: high retentivity.
2. It must also resist demagnetisation by stray fields: high coercivity.
3. So both retentivity and coercivity should be high (steel is a typical choice).

Teacher's Note:
1. Retentivity measures how much magnetism remains.
2. Coercivity measures the reverse field needed to remove it.

Answer: (c) \( 10^{-3} \) N

Explanation:
1. The horizontal field of the earth points north-south, perpendicular to the east-west wire.
2. Force per unit length \( = BI = 10^{-4}\times 10 \).
3. So \( F/L = 10^{-3} \) N per metre.

Teacher's Note:
1. Use \( F = BIL\sin\theta \) with \( \theta = 90^\circ \).
2. Check that the wire is perpendicular to the field.

Answer: (a) \( \frac{17}{10\pi} \) A

Explanation:
1. The coil's field at the centre, \( B = \frac{\mu_0 N i}{2r} \), is perpendicular to the earth's horizontal field \( B_H \) (the coil is in the meridian).
2. For deflection \( \theta \): \( B = B_H\tan\theta = 0.34\times 10^{-4} \) T at \( 45^\circ \).
3. \( i = \frac{B_H\tan 45^\circ\times 2r}{\mu_0 N} = \frac{0.34\times 10^{-4}\times 0.4}{4\pi\times 10^{-7}\times 20} = \frac{17}{10\pi} \) A (about 0.54 A).

Teacher's Note:
1. Tangent law: \( B = B_H\tan\theta \).
2. The coil's axis is east-west, perpendicular to the meridian.

Answer: (a) 0.1112 A and 0.0963 A

Explanation:
1. The vertical coil has a horizontal axis, so it cancels the horizontal component: \( \frac{\mu_0 N I_V}{2R_V} = B_H \), giving \( I_V = \frac{2\times 0.2\times 27.8}{100} = 0.1112 \) A.
2. The horizontal coil has a vertical axis, so it cancels the vertical component \( B_V = B_H\tan 30^\circ = \frac{B_H}{\sqrt{3}} \).
3. \( I_H = \frac{2R_H B_V}{\mu_0 N} = \frac{2\times 0.3\times 27.8}{\sqrt{3}\times 100} \approx 0.0963 \) A.

Teacher's Note:
1. Each coil's field is along its own axis.
2. \( B_V = B_H\tan(\text{dip}) \).

Answer: (a) \( 1.67\times 10^{-4} \) T

Explanation:
1. Axial field of the magnet: \( B_1 = \frac{\mu_0}{4\pi}\frac{2M}{r^3} = 10^{-7}\times\frac{2\times 6}{0.008} = 1.5\times 10^{-4} \) T, along the magnet's axis (\( 60^\circ \) east of north).
2. The earth's horizontal field is \( B_2 = 0.3\times 10^{-4} \) T along north, at \( 60^\circ \) to \( B_1 \).
3. \( B = \sqrt{B_1^2 + B_2^2 + 2B_1B_2\cos 60^\circ} = 10^{-4}\sqrt{2.25 + 0.09 + 0.45} \approx 1.67\times 10^{-4} \) T.

Teacher's Note:
1. Add the two fields as vectors using the parallelogram law.
2. The axial field formula is \( \frac{\mu_0}{4\pi}\frac{2M}{r^3} \).

Answer: (a) 2.0 V

Explanation:
1. Wire length \( = 50\times 2\pi\times 0.1 = 31.4 \) m and area \( = \pi(0.1\times 10^{-3})^2 \), so \( R = \frac{\rho L}{A} = \frac{2\times 10^{-8}\times 31.4}{3.14\times 10^{-8}} = 20\ \Omega \).
2. To cancel \( B_H \): \( \frac{\mu_0 N i}{2r} = 3.14\times 10^{-5} \), so \( i = \frac{3.14\times 10^{-5}\times 0.2}{4\pi\times 10^{-7}\times 50} = 0.1 \) A.
3. \( V = iR = 0.1\times 20 = 2.0 \) V (the coil must be vertical, perpendicular to the magnetic meridian).

Teacher's Note:
1. Convert the resistivity to ohm metre before computing \( R \).
2. Use the field at the centre of a circular coil.

Answer: (a) 596

Explanation:
1. \( B = \frac{\Phi}{A} = \frac{2.4\times 10^{-5}}{0.2\times 10^{-4}} = 1.2 \) T.
2. Permeability \( \mu = \frac{B}{H} = \frac{1.2}{1600} = 7.5\times 10^{-4} \) N A\( ^{-2} \).
3. \( \chi_m = \frac{\mu}{\mu_0} - 1 = \frac{7.5\times 10^{-4}}{4\pi\times 10^{-7}} - 1 \approx 596 \).

Teacher's Note:
1. Convert the area to square metres.
2. Use \( \mu = \mu_0(1 + \chi_m) \).

Answer: (a) 799

Explanation:
1. The mean radius is \( 11.5 \) cm, so the turns per unit length are \( n = \frac{3000}{2\pi\times 0.115} \).
2. With the core: \( B = \mu_0\mu_r n i \), so \( \mu_r = \frac{B}{\mu_0 n i} \).
3. \( \mu_r = \frac{2.5\times 0.115}{2\times 10^{-7}\times 3000\times 0.6} \approx 799 \).

Teacher's Note:
1. Use the mean radius to find the turns per unit length.
2. The same field without the core would be \( \mu_0 n i \).