JEE Physics Magnetism And Matter MCQs Set 04

Answer: (b) \( \sqrt5\times10^{-7}\ \text{T} \)

Explanation:
1. Each dipole is 1 m from the midpoint. Axial field: \( B_a=\frac{\mu_0}{4\pi}\frac{2M}{r^3}=2\times10^{-7}\ \text{T} \). Equatorial field: \( B_e=\frac{\mu_0}{4\pi}\frac{M}{r^3}=10^{-7}\ \text{T} \).
2. The two fields are perpendicular: \( B=\sqrt{(2)^2+(1)^2}\times10^{-7}=\sqrt5\times10^{-7}\ \text{T} \).

Teacher's Note:
1. Axial field is twice the equatorial field for the same distance.
2. Add perpendicular fields by Pythagoras' theorem.

Answer: (b) 4 : 1

Explanation:
1. At O, M1 is on its axial line (distance 2d): \( B_1=\frac{\mu_0}{4\pi}\frac{2M_1}{(2d)^3} \). M2 is on its equatorial line (distance d): \( B_2=\frac{\mu_0}{4\pi}\frac{M_2}{d^3} \).
2. For zero resultant \( B_1=B_2 \): \( \frac{2M_1}{8}=M_2\Rightarrow\frac{M_1}{M_2}=4 \).

Teacher's Note:
1. Axial field is 2kM/r³ and equatorial field is kM/r³.
2. The fields cancel when their magnitudes are equal.

Answer: (a) \( \frac{\mu}{1.4} \)

Explanation:
1. Pole strength is unchanged on breaking, so the moments are proportional to the lengths: \( \mu_1=\frac{6}{14}\mu,\ \mu_2=\frac{8}{14}\mu \).
2. At right angles: \( \mu_{net}=\sqrt{\mu_1^2+\mu_2^2}=\mu\frac{\sqrt{36+64}}{14}=\frac{10}{14}\mu=\frac\mu{1.4} \).

Teacher's Note:
1. Moment = pole strength × length.
2. Perpendicular vectors add by Pythagoras.

Answer: (d) \( \frac{\tau_1}{\tau_2} \)

Explanation:
1. Perpendicular to the field, \( \tau=MB\sin90^\circ=MB \).
2. So \( \frac{B_1}{B_2}=\frac{\tau_1}{\tau_2} \).

Teacher's Note:
1. The same needle has the same magnetic moment at both places.
2. Torque is proportional to the field.

Answer: (c) \( \frac{3M}{\pi} \)

Explanation:
1. Let the wire have length L and pole strength m, so \( M=mL \).
2. Arc subtends 60° (\( \pi/3 \)): radius \( R=\frac{L}{\pi/3}=\frac{3L}{\pi} \).
3. The new moment is \( m \) times the distance between the poles (the chord): chord \( =2R\sin30^\circ=R=\frac{3L}{\pi} \), so \( M'=\frac{3mL}{\pi}=\frac{3M}{\pi} \).

Teacher's Note:
1. Magnetic moment is pole strength times the distance between the poles.
2. The chord of a 60° arc equals the radius.

Answer: (c) both spin and orbital motion

Explanation:
1. An electron has an orbital magnetic moment due to its motion around the nucleus.
2. It also has a spin magnetic moment due to its intrinsic spin.
3. The total magnetic field arises from both.

Teacher's Note:
1. Both orbital and spin contribute to the atomic magnetic moment.
2. In many atoms the contributions of paired electrons cancel.

Answer: (b) always decreases

Explanation:
1. Heating increases the thermal agitation of the atomic dipoles, which disturbs their alignment.
2. So the magnetisation, and hence the strength, decreases (it vanishes at the Curie temperature).

Teacher's Note:
1. Thermal motion randomises the magnetic domains.
2. Above the Curie temperature a ferromagnet becomes paramagnetic.

Answer: (a) parallel and straight

Explanation:
1. Over a small region the earth's horizontal component is uniform in magnitude and direction.
2. Uniform field lines are parallel straight lines.

Teacher's Note:
1. A uniform field is represented by equally spaced parallel lines.
2. Over a small region the earth's field is uniform.

Answer: (d) the horizontal component of the earth's field

Explanation:
1. A deflection magnetometer balances the field of a bar magnet against the horizontal component of the earth's field \( B_H \).
2. Using the deflection angle, \( B=B_H\tan\theta \), it gives the ratio of the magnet's field to \( B_H \) and hence \( B_H \).

Teacher's Note:
1. The tangent law connects the deflection to B_H.
2. Dip needs a dip circle; declination needs a compass and a meridian.

Answer: (c) agonic lines

Explanation:
1. Isogonic lines join places of equal declination, isoclinic lines join places of equal dip, and aclinic lines join places of zero dip (the magnetic equator).
2. Lines joining places of zero declination are agonic lines.

Teacher's Note:
1. Agonic means 'no angle'.
2. Aclinic lines have zero dip.

Answer: (b) diamagnetic

Explanation:
1. A hydrogen atom has one unpaired electron, so it is paramagnetic.
2. In the H₂ molecule the two electrons are paired with opposite spins, so the net magnetic moment is zero: it is diamagnetic.

Teacher's Note:
1. Paired electrons give no net moment.
2. A diamagnetic substance has all its electrons paired.

Answer: (c) both of these

Explanation:
1. High susceptibility lets the core become strongly magnetised by a small current.
2. Low coercivity lets it lose the magnetism easily when the current is switched off.
3. Both properties are required, so both statements are correct.

Teacher's Note:
1. An electromagnet must switch on strongly and off quickly.
2. Soft iron has a thin, narrow hysteresis loop.

Answer: (a) attracts some substances and repels others

Explanation:
1. A magnet attracts ferromagnetic and paramagnetic substances.
2. It weakly repels diamagnetic substances.
3. So it attracts some substances and repels others.

Teacher's Note:
1. Paramagnetic substances are weakly attracted; diamagnetic substances are weakly repelled.
2. The effect is strong only for ferromagnetic materials.

Answer: (d) X is diamagnetic and Y is paramagnetic

Explanation:
1. Diamagnetic: \( \mu_r \) slightly less than 1.
2. Paramagnetic: \( \mu_r \) slightly more than 1.
3. Ferromagnetic: \( \mu_r \) much greater than 1.
4. So X is diamagnetic and Y is paramagnetic.

Teacher's Note:
1. μr = 1 + χ, with χ small and negative for diamagnetic materials.
2. Ferromagnets have very large μr.

Answer: (d) negative and independent of temperature

Explanation:
1. For a diamagnetic material the induced magnetisation opposes the field, so χ is negative.
2. It arises from the orbital motion of electrons and does not depend on temperature.

Teacher's Note:
1. Curie's law applies only to paramagnetic materials.
2. Diamagnetism is present in all materials but is weak.