JEE Physics Magnetism and Matter MCQs Set 27

Practice MCQs for JEE Physics Magnetism and Matter

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Multiple Choice Questions

Question 1:
Assertion (A): The flux of the magnetic field \( \vec B \) through a closed surface is equal to zero, \( \oint\vec B\cdot d\vec s=0 \).
Reason (R): Magnetic field lines are closed curves; they do not have any beginning or end.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Every field line that enters a closed surface also leaves it, so the net flux is zero.
2. This is because the lines form closed loops, so R explains A.

Teacher's Note:
1. There are no isolated magnetic monopoles.
2. This is Gauss' law for magnetism.

Question 2: A magnet of length 14 cm and magnetic moment \( M \) is broken into two parts of lengths 6 cm and 8 cm. They are put at right angles to each other with opposite poles together. The magnetic moment of the combination is
(a) \( M/10 \)
(b) \( M \)
(c) \( M/1.4 \)
(d) \( 2.8M \)
Show Answer & Explanation

Answer: (c) \( M/1.4 \)

Explanation:
1. Pole strength \( m \) is unchanged: \( M=14m \), and the parts have moments \( 6m \) and \( 8m \).
2. The moments are perpendicular: \( \sqrt{(6m)^2+(8m)^2}=10m=\dfrac{10}{14}M=\dfrac{M}{1.4} \).

Teacher's Note:
1. Breaking a magnet does not change its pole strength, only its length.

Question 3: The needle of a deflection galvanometer shows a deflection of \( 60^\circ \) due to a short bar magnet at a certain distance in tan A position. If the distance is doubled, the deflection becomes
(a) \( \sin^{-1}\left[\dfrac{\sqrt3}{8}\right] \)
(b) \( \cos^{-1}\left[\dfrac{\sqrt3}{8}\right] \)
(c) \( \tan^{-1}\left[\dfrac{\sqrt3}{8}\right] \)
(d) \( \cot^{-1}\left[\dfrac{\sqrt3}{8}\right] \)
Show Answer & Explanation

Answer: (c) \( \tan^{-1}\left[\dfrac{\sqrt3}{8}\right] \)

Explanation:
1. \( \tan\theta=\dfrac{B}{B_H}\propto\dfrac1{d^3} \).
2. Doubling \( d \) divides \( \tan\theta \) by 8: \( \tan\theta'=\dfrac{\tan60^\circ}{8}=\dfrac{\sqrt3}{8} \).

Teacher's Note:
1. The field of a short magnet falls as the cube of the distance.

Question 4: A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of the combination is 4 s. If one of the magnets is removed, the period of the other is
(a) 5 s
(b) 3.36 s
(c) 4.36 s
(d) 5.36 s
Show Answer & Explanation

Answer: (b) 3.36 s

Explanation:
1. Combination: \( M'=\sqrt2M \), \( I'=2I \).
2. \( T'=2\pi\sqrt{\dfrac{2I}{\sqrt2MB}}=2^{1/4}\,T \), so \( T=\dfrac{4}{2^{1/4}}=3.36 \) s.

Teacher's Note:
1. Net moment of perpendicular magnets is \( \sqrt2M \) while the moment of inertia doubles.

Question 5: A magnetic needle lying parallel to a magnetic field requires \( W \) units of work to turn it through \( 60^\circ \). The torque required to keep the needle in this position will be
(a) \( 2W \)
(b) \( W \)
(c) \( W/\sqrt3 \)
(d) \( \sqrt3W \)
Show Answer & Explanation

Answer: (d) \( \sqrt3W \)

Explanation:
1. \( W=MB(1-\cos60^\circ)=\dfrac{MB}2\Rightarrow MB=2W \).
2. \( \tau=MB\sin60^\circ=2W\times\dfrac{\sqrt3}2=\sqrt3W \).

Teacher's Note:
1. Work uses \( 1-\cos\theta \); torque uses \( \sin\theta \).

Question 6: Two identical magnetic dipoles of magnetic moment 2 A m\(^2\) each are placed at a separation of 2 m with their axes perpendicular to each other and bisecting each other in air. The resultant magnetic field at the mid point between the dipoles is
(a) \( 4\sqrt5\times10^{-5} \) T
(b) \( 2\sqrt5\times10^{-5} \) T
(c) \( 4\sqrt5\times10^{-7} \) T
(d) \( 2\sqrt5\times10^{-7} \) T
Show Answer & Explanation

Answer: (d) \( 2\sqrt5\times10^{-7} \) T

Explanation:
1. The mid point is 1 m from each dipole; it lies on the axial line of one and on the equatorial line of the other.
2. \( B_1=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}=4\times10^{-7} \) T and \( B_2=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}=2\times10^{-7} \) T, at right angles.
3. \( B=\sqrt{16+4}\times10^{-7}=2\sqrt5\times10^{-7} \) T.

Teacher's Note:
1. Axial field is twice the equatorial field for the same \( M \) and \( r \).

Question 7: The magnetic field due to a short bar magnet of magnetic dipole moment \( M \) and length \( 2l \), on the axis at a distance \( z \) (\( z\gg l \)) from the centre of the magnet is given by
(a) \( \dfrac{\mu_0M}{4\pi z^3}\hat M \)
(b) \( \dfrac{2\mu_0M}{4\pi z^3}\hat M \)
(c) \( \dfrac{4\pi M}{\mu_0z^2}\hat M \)
(d) \( \dfrac{\mu_0M}{2\pi z^3}\hat M \)
Show Answer & Explanation

Answer: (b) \( \dfrac{2\mu_0M}{4\pi z^3}\hat M \)

Explanation:
1. The axial field of a short dipole is \( \dfrac{\mu_0}{4\pi}\dfrac{2M}{z^3} \), directed along \( \vec M \).

Teacher's Note:
1. Axial: \( 2M/z^3 \); equatorial: \( M/r^3 \) (opposite to \( \vec M \)).

Question 8: A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction \( B \). If it is rotated through \( 180^\circ \), the work done is (\( M \) = magnetic dipole moment)
(a) \( MB \)
(b) \( 2MB \)
(c) \( MB/2 \)
(d) zero
Show Answer & Explanation

Answer: (b) \( 2MB \)

Explanation:
1. \( W=MB(\cos0^\circ-\cos180^\circ)=2MB \).

Teacher's Note:
1. Stable equilibrium has energy \( -MB \); unstable has \( +MB \).

Question 9: A bar magnet is cut into four equal parts such that the length and breadth of each part are equal. The pole strength of each part is (original pole strength \( m \))
(a) \( m \)
(b) \( m/2 \)
(c) \( m/4 \)
(d) \( m/8 \)
Show Answer & Explanation

Answer: (c) \( m/4 \)

Explanation:
1. Pole strength is proportional to the cross-sectional area.
2. Each part has one quarter of the area, so its pole strength is \( m/4 \).

Teacher's Note:
1. Cutting across the length keeps pole strength; cutting along it divides it.

Question 10: A magnetic dipole is placed in a uniform magnetic field. The net magnetic force on the dipole
(a) is always zero
(b) depends on the orientation of the dipole
(c) can never be zero
(d) depends on the strength of the dipole
Show Answer & Explanation

Answer: (a) is always zero

Explanation:
1. The forces on the two poles are equal and opposite in a uniform field, so the net force is zero.
2. Only a torque acts, unless the dipole is aligned.

Teacher's Note:
1. A non-uniform field is needed for a net force.

Question 11: A short bar magnet has a length \( 2l \) and a magnetic moment 10 A m\(^2\). The magnetic field at a distance \( r=0.1 \) m from its centre on the axial line is (\( l \) is negligible compared to \( r \))
(a) \( 2\times10^{-3} \) T
(b) \( 3\times10^{-3} \) T
(c) \( 1\times10^{-3} \) T
(d) \( 4\times10^{-3} \) T
Show Answer & Explanation

Answer: (a) \( 2\times10^{-3} \) T

Explanation:
1. \( B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}=10^{-7}\times\dfrac{20}{10^{-3}}=2\times10^{-3} \) T.

Teacher's Note:
1. \( \mu_0/4\pi=10^{-7} \) T m/A.

Question 12: A short bar magnet experiences a torque of magnitude 0.64 J when it is placed in a uniform magnetic field of 0.32 T, making an angle of \( 30^\circ \) with the field. The magnetic moment of the magnet is
(a) 1 A m\(^2\)
(b) 4 A m\(^2\)
(c) 6 A m\(^2\)
(d) none of these
Show Answer & Explanation

Answer: (b) 4 A m\(^2\)

Explanation:
1. \( \tau=MB\sin\theta\Rightarrow M=\dfrac{0.64}{0.32\times0.5}=4 \) A m\(^2\).

Teacher's Note:
1. Use \( \sin30^\circ=\tfrac12 \).

Question 13: The magnetic moment of a bar magnet is \( M \). The work done to turn the magnet through \( 90^\circ \) from the direction of the magnetic field \( B \) is
(a) zero
(b) \( \tfrac12MB \)
(c) \( 2MB \)
(d) \( MB \)
Show Answer & Explanation

Answer: (d) \( MB \)

Explanation:
1. \( W=MB(\cos0^\circ-\cos90^\circ)=MB \).

Teacher's Note:
1. Work is \( MB(1-\cos\theta) \).

Question 14: The magnetic field of a small bar magnet at a large distance \( d \) varies as
(a) \( 1/d \)
(b) \( 1/d^2 \)
(c) \( 1/d^3 \)
(d) \( 1/d^4 \)
Show Answer & Explanation

Answer: (c) \( 1/d^3 \)

Explanation:
1. The field of a magnetic dipole falls as \( 1/d^3 \).

Teacher's Note:
1. Electric dipoles show the same \( 1/r^3 \) behaviour.

Question 15: An electron of charge \( e \) moves in a circular orbit of radius \( r \) around the nucleus at a frequency \( \nu \). The magnetic moment associated with the orbital motion of the electron is
(a) \( \pi\nu er^2 \)
(b) \( \dfrac{\pi\nu r^2}{e} \)
(c) \( \dfrac{\pi\nu e}{r} \)
(d) \( \dfrac{\pi er^2}{\nu} \)
Show Answer & Explanation

Answer: (a) \( \pi\nu er^2 \)

Explanation:
1. The orbital current is \( I=e\nu \).
2. \( M=IA=e\nu\cdot\pi r^2 \).

Teacher's Note:
1. Moment of a current loop is current times area.

Question 16: A bar magnet when placed at an angle of \( 30^\circ \) to the direction of a magnetic induction of \( 5\times10^{-2} \) T experiences a moment of couple \( 25\times10^{-6} \) N m. If the length of the magnet is 5 cm, its pole strength is
(a) \( 2\times10^{-2} \) A m
(b) \( 5\times10^{-2} \) A m
(c) 2 A m
(d) 5 A m
Show Answer & Explanation

Answer: (a) \( 2\times10^{-2} \) A m

Explanation:
1. \( M=\dfrac{\tau}{B\sin\theta}=\dfrac{25\times10^{-6}}{5\times10^{-2}\times0.5}=10^{-3} \) A m\(^2\).
2. \( m=\dfrac Ml=\dfrac{10^{-3}}{0.05}=2\times10^{-2} \) A m.

Teacher's Note:
1. \( M=m\times\) (magnetic length).

Question 17: A current carrying small loop behaves like a small magnet. If \( A \) is its area and \( M \) its magnetic moment, the current in the loop is
(a) \( M/A \)
(b) \( A/M \)
(c) \( MA \)
(d) \( AM^2 \)
Show Answer & Explanation

Answer: (a) \( M/A \)

Explanation:
1. \( M=IA\Rightarrow I=M/A \).

Teacher's Note:
1. A loop of current is a magnetic dipole.

Question 18: A magnet of magnetic moment 20 CGS units is freely suspended in a uniform magnetic field of intensity 0.3 CGS units. The amount of work done in deflecting it by an angle of \( 30^\circ \) in CGS units is
(a) 6
(b) \( 3\sqrt3 \)
(c) \( 3(2-\sqrt3) \)
(d) 3
Show Answer & Explanation

Answer: (c) \( 3(2-\sqrt3) \)

Explanation:
1. \( W=MH(1-\cos30^\circ)=20\times0.3\left(1-\dfrac{\sqrt3}2\right)=3(2-\sqrt3) \).

Teacher's Note:
1. Start from the equilibrium position, \( \theta=0 \).

Question 19: At a point on the right bisector of a magnetic dipole, the magnetic
(a) potential varies as \( 1/r^2 \)
(b) potential is zero at all points on the right bisector
(c) field varies as \( r^2 \)
(d) field is perpendicular to the axis of the dipole
Show Answer & Explanation

Answer: (b) potential is zero at all points on the right bisector

Explanation:
1. Magnetic scalar potential \( V=\dfrac{\mu_0}{4\pi}\dfrac{M\cos\theta}{r^2} \) and \( \theta=90^\circ \) on the bisector, so \( V=0 \).
2. The equatorial field varies as \( 1/r^3 \) and is antiparallel to the axis.

Teacher's Note:
1. On the equatorial line the potential vanishes but the field does not.

Question 20: The magnetism of a magnet is due to
(a) the earth
(b) cosmic rays
(c) the spin motion of electrons
(d) pressure of big magnet inside the earth
Show Answer & Explanation

Answer: (c) the spin motion of electrons

Explanation:
1. Magnetism arises from the spin and orbital motion of electrons, mainly their spin.

Teacher's Note:
1. Atomic dipoles add up in ferromagnets.

Question 21: The magnetic lines of force inside a bar magnet
(a) are from north pole to south pole of the magnet
(b) do not exist
(c) depend upon the area of cross-section of the bar magnet
(d) are from south pole to north pole of the magnet
Show Answer & Explanation

Answer: (d) are from south pole to north pole of the magnet

Explanation:
1. Field lines are closed loops: outside they go from N to S, so inside the magnet they run from S to N.

Teacher's Note:
1. Closed loops mean the lines continue inside the magnet.

Question 22: The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows
(a) both torque and net force are present
(b) torque is present but no net force
(c) both torque and net force are absent
(d) net force is present but no torque
Show Answer & Explanation

Answer: (b) torque is present but no net force

Explanation:
1. In a uniform field the two poles feel equal and opposite forces: zero net force but a torque \( MB\sin\theta \).

Teacher's Note:
1. The torque aligns the needle with the field.

Question 23: A bar magnet is held at right angles to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it from this position. The angle of rotation is
(a) \( 60^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 75^\circ \)
Show Answer & Explanation

Answer: (a) \( 60^\circ \)

Explanation:
1. Initial torque \( MB\sin90^\circ=MB \); we need \( MB\sin\theta=MB/2 \), so \( \theta=30^\circ \).
2. Rotation \( =90^\circ-30^\circ=60^\circ \).

Teacher's Note:
1. Take care: the angle asked is the rotation, not the final angle.

Question 24: A magnetic needle is kept in a non-uniform magnetic field. It experiences
(a) a torque but not a force
(b) neither a force nor a torque
(c) a force and a torque
(d) a force but not a torque
Show Answer & Explanation

Answer: (c) a force and a torque

Explanation:
1. In a non-uniform field the poles feel unequal forces, giving a net force as well as a torque.

Teacher's Note:
1. Compare with a uniform field, where only torque acts.

Question 25: The direction of the magnetic field at an equatorial point of a short magnetic dipole of moment \( \vec M \) is
(a) parallel to \( \vec M \)
(b) perpendicular to \( \vec M \)
(c) making an angle of \( 45^\circ \) with \( \vec M \)
(d) anti-parallel to \( \vec M \)
Show Answer & Explanation

Answer: (d) anti-parallel to \( \vec M \)

Explanation:
1. On the equatorial line \( \vec B=-\dfrac{\mu_0}{4\pi}\dfrac{\vec M}{r^3} \), opposite to \( \vec M \).

Teacher's Note:
1. Axial field is along \( \vec M \); equatorial field is against it.

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