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Physics MCQs for NEET — Practice Questions with Answers

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Points A and B are situated perpendicular to the axis of a 2cm long bar magnet at large distances X and 3X from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to

1 : 9

2 : 9

27 : 1

9 : 1

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Two short magnets with their axes horizontal to the magnetic meridian are placed with their centres 40 cm east and 50 cm west of the magnetic needle. If the needle remains undeflected, the ratio of their magnetic moments is

4:5

16:25

64:125

2: $\sqrt{5}$

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Explanation

(3)

$M a g n e t i c f i e l d d u e t o a s h o r t b a r m a g n e t o n i t s e q u i t o r i a l l i n e = \frac{μ_{0}}{4 π} (\frac{M}{d^{3}}) A s t h e r e i s n o d e f l e c t i o n i n m a g n e t i c n e e d l e, B_{1} = B_{2} \Rightarrow \frac{μ_{0}}{4 π} (\frac{M_{1}}{{d_{1}}^{3}}) = \frac{μ_{0}}{4 π} (\frac{M_{2}}{{d_{2}}^{3}})$

$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3} = {(\frac{40}{50})}^{3} = \frac{64}{125}$

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If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B, the work done in rotating the magnet through an angle is

$M B (1 - \sin θ)$

$M B \sin θ$

$M B \cos θ$

$M B (1 - \cos θ)$

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Explanation

(d)

$W = M B (\cos θ_{1} - \cos θ) = M B (\cos 0 - \cos θ) = M B (1 - \cos θ)$

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Two small bar magnets are placed in a line with like poles facing each other at a certain distance d apart. If the length of each magnet is negligible as compared to d, the force between them will be inversely proportional to

d²

$\frac{1}{d^{2}}$

$d^{4}$

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A magnet of magnetic moment M is situated with its axis along the direction of a magnetic field of strength B. The work done in rotating it by an angle of 180^o will be

-MB

+MB

+2MB

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A magnetic needle is kept in a non-uniform magnetic field. It experiences

A force and a torque

A force but not a torque

A torque but not a force

Neither a torque nor a force

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Explanation

(a)

In a non-uniform magnetic field, the force on each of the poles will be different in both magnitude and direction.
Due to difference in magnitude, the dipole experiences a force, Due to differences in direction, the dipole experiences a torque.

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A long magnetic needle of length 2L, magnetic moment M and pole strength m units is broken into two pieces at the middle. The magnetic moment and pole strength of each piece will be

$\frac{M}{2}, \frac{m}{2}$

$M, \frac{m}{2}$

$\frac{M}{2}, m$

M, m

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Two identical thin bar magnets each of length l and pole strength m are placed at the right angle to each other with the north pole of one touching south pole of the other. The magnetic moment of the system is :

2ml

$\sqrt{2} m l$

$\frac{1}{2} m l$

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Explanation

When two identical bar magnets of length l and pole strength m are placed at right angles with their poles touching, their magnetic moments form a resultant at an angle of 45Â° to each magnet. The magnitude of the resultant magnetic moment is âˆš2ml, as confirmed by the vector addition of the two equal magnetic moments ml.

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Rate of change of torque with deflections maximum for a magnet suspended freely in a uniform magnetic field of induction B, when

$θ = 0^{o}$

$θ = 45^{o}$

$θ = 60^{o}$

$θ = 90^{o}$

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A magnet of magnetic moment $50 \hat{i} A - m^{2}$ is placed along the x-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) T$ . The torque acting on the magnet is

175 $\hat{k}$ N-m

150 $\hat{k}$ N-m

75 $\hat{k}$ N-m

25 $\sqrt{37}$ $\hat{k}$ N-m

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Explanation

The torque acting on a magnetic dipole moment (m) in a uniform magnetic field (B) is given by Ï„ = m Ã— B. Here, m = 50 Am^2 along x-axis and B = (0.5i + 3j) T. Taking the cross product, we get Ï„ = (50)(3) = 150k N-m.

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