JEE Physics Magnetism and Matter MCQs Set 03

Answer: (b) 2

Explanation:
1. \( W=MB(1-\cos\theta) \).
2. \( W_{90}=MB \), \( W_{60}=MB(1-\tfrac12)=\tfrac{MB}2 \).
3. So n = 2.

Teacher's Note:
1. The work is measured from the equilibrium position θ = 0.
2. cos 60° = 1/2.

Answer: (d) never intersect each other

Explanation:
1. At every point the field has one definite direction, which is the tangent to the line.
2. If two lines crossed, the field would have two directions at the crossing point, which is impossible.

Teacher's Note:
1. Neutral points are where the field is zero, not where lines cross.
2. The same holds for electric field lines.

Answer: (a) \( \sqrt3:\sqrt2 \)

Explanation:
1. \( B_H=B\cos\delta \).
2. \( \frac{B_{H1}}{B_{H2}}=\frac{\cos30^\circ}{\cos45^\circ}=\frac{\sqrt3/2}{1/\sqrt2}=\sqrt{\tfrac32} \)=\( \sqrt3:\sqrt2 \).

Teacher's Note:
1. The horizontal component is B cos δ, the vertical is B sin δ.
2. Keep the ratio in the order stated.

Answer: (a) \( T_1<T_2 \)

Explanation:
1. \( T=2\pi\sqrt{I/M_{net}B} \).
2. Same sense: \( M_{net}=3M \). Reversed: \( M_{net}=M \).
3. A smaller net moment gives a longer period, so T1 < T2.

Teacher's Note:
1. Net moment is the vector sum of the two.
2. T2 is not infinite since the net moment is M, not zero.

Answer: (d) 1.2 G

Explanation:
1. \( f\propto\sqrt B \).
2. Doubling f needs B to be 4 times, i.e. 1.2 G.

Teacher's Note:
1. Frequency varies as the square root of the field.
2. The question asks for the new field value.

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

Explanation:
1. Second place: \( f_2=\frac{1}{2.5}=0.4 \) Hz = 24 per minute.
2. \( \frac{B_2}{B_1}=\left(\frac{24}{40}\right)^2=0.36 \).
3. \( B_2=0.36\times10^{-6} \) T.

Teacher's Note:
1. Convert both to the same time unit.
2. B ∝ f².

Answer: (b) \( 1.96\times10^{-5} \) T

Explanation:
1. Tangent law: \( B_{coil}=B_H\tan\theta \).
2. \( =0.34\times10^{-4}\times0.577\approx1.96\times10^{-5} \) T.

Teacher's Note:
1. The coil field is perpendicular to the earth's horizontal component.
2. tan 30° = 0.577.

Answer: (a) an iron box

Explanation:
1. Iron has a very high permeability, so field lines prefer to pass through its walls.
2. The space inside is shielded; wood and non-magnetic metals do not shield.

Teacher's Note:
1. This is magnetic shielding by a soft-iron ring or box.
2. Aluminium or copper cannot shield a static field.

Answer: (d) copper

Explanation:
1. Iron, cobalt and steel are ferromagnetic and are attracted.
2. Copper is diamagnetic and is weakly repelled.

Teacher's Note:
1. Diamagnets have negative susceptibility.
2. Examples: copper, bismuth, water, gold.

Answer: (a) slightly less

Explanation:
1. \( B=\mu_0\mu_rH \) and \( \mu_r<1 \) for a diamagnet.
2. So the flux density inside is slightly smaller.

Teacher's Note:
1. Diamagnets expel the field slightly.
2. Ferromagnets concentrate the field.

Answer: (d) ferromagnetic substance

Explanation:
1. A ferromagnetic container has very high permeability and carries the field lines around the interior.
2. Diamagnetic and paramagnetic materials are too weak to shield.

Teacher's Note:
1. Soft iron is the usual shielding material.
2. Shielding works by diverting field lines.

Answer: (d) both ferromagnetic and paramagnetic

Explanation:
1. In diamagnetic atoms the moments cancel exactly, so the net atomic moment is zero.
2. Paramagnetic and ferromagnetic atoms carry a permanent atomic moment (unpaired electrons).

Teacher's Note:
1. The difference between para and ferro is in the interaction between neighbouring moments.
2. Diamagnetism is a universal weak induced effect.

Answer: (b) hysteresis

Explanation:
1. Paramagnetic materials are also attracted and have no hysteresis loop.
2. Only ferromagnets show a non-reversible B-H loop (hysteresis).

Teacher's Note:
1. Hysteresis comes from domain wall motion.
2. Susceptibility of para- and ferromagnets depends on temperature.

Answer: (b) a broad hysteresis loop

Explanation:
1. Hard materials have high coercivity and high retentivity.
2. This gives a broad hysteresis loop, ideal for permanent magnets.

Teacher's Note:
1. Magnetic hardness is not mechanical hardness.
2. Soft materials have narrow loops.

Answer: (c) equal to zero

Explanation:
1. In a diamagnetic atom the orbital and spin moments cancel in pairs.
2. The net permanent moment is zero.

Teacher's Note:
1. Diamagnetism arises from induced moments only.
2. The induced moment opposes the applied field.

Answer: (d) all of the above

Explanation:
1. \( \mu_r=1+\chi \) with \( \chi \) small and negative, so \( \mu_r<1 \).
2. Diamagnetism is independent of temperature.

Teacher's Note:
1. All three statements are standard properties.
2. Check each one individually.

Answer: (b) less

Explanation:
1. The loop area is the energy loss per cycle per unit volume.
2. Soft iron has low coercivity, so its loop is narrow and its area smaller than steel's.

Teacher's Note:
1. Small loop area means small hysteresis loss.
2. That is why soft iron suits transformer cores.

Answer: (d) cobalt-steel

Explanation:
1. Permanent magnets need high coercivity and high retentivity.
2. Alloys such as cobalt steel, alnico and Nipermag are good choices.

Teacher's Note:
1. Pure cobalt has lower coercivity than its steel alloys.
2. Aluminium is paramagnetic.

Answer: (b) It becomes paramagnetic at the Curie temperature

Explanation:
1. Above the Curie temperature thermal agitation destroys domain alignment.
2. The material behaves as a paramagnet, following the Curie-Weiss law.

Teacher's Note:
1. Susceptibility does not become zero; it falls as 1/(T − Tc).
2. Iron's Curie point is about 1043 K.

Answer: (a) diamagnetic

Explanation:
1. Para- and ferromagnetic rods align parallel to the field (to regions of stronger field).
2. A rod that lines up across the field is diamagnetic.

Teacher's Note:
1. Diamagnets move to weaker field regions.
2. A rod of copper or bismuth turns across the field.

Answer: (d) 1

Explanation:
1. Air is only very feebly paramagnetic, with \( \chi\approx4\times10^{-7} \).
2. So \( \mu_r=1+\chi\approx1 \).

Teacher's Note:
1. Nonmagnetic media have μ_r equal to 1 in practice.
2. Vacuum has exactly 1.

Answer: (a) diamagnetic

Explanation:
1. A small negative susceptibility characterises diamagnetism.
2. Paramagnets have small positive values; ferromagnets large positive values.

Teacher's Note:
1. Sign of χ tells the nature.
2. Typical diamagnetic χ is about −10⁻⁵ to −10⁻³.

Answer: (c) coercivity

Explanation:
1. Coercivity is the reverse field needed to bring the magnetisation to zero.
2. Hard materials have large coercivity; they resist demagnetisation.

Teacher's Note:
1. Coercivity, not retentivity, defines hardness.
2. Retentivity and coercivity are both high for permanent magnets.

Answer: (b) diamagnetic

Explanation:
1. \( \chi=\mu_r-1=-10^{-4}<0 \).
2. Negative χ means diamagnetic.

Teacher's Note:
1. μ_r slightly below 1 signals diamagnetism.
2. μ_r slightly above 1 signals paramagnetism.

Answer: (c) 0.015

Explanation:
1. Curie's law: \( \chi\propto\frac1T \).
2. T1 = 200 K, T2 = 100 K. So \( \chi_2=0.0075\times\frac{200}{100}=0.015 \).

Teacher's Note:
1. Convert °C to kelvin before applying Curie's law.
2. Cooling a paramagnet raises its susceptibility.