NEET Physics Magnetism and Matter Questions Solved

A bar magnet of length l and magnetic dipole moment M is bent to form an arc which subtends an angle of $120 °$ at centre. The new magnetic dipole moment will be

1. $\frac{3 M}{2 π}$

2. $\frac{3 \sqrt{3} M}{2 π}$

3. $\frac{3 M}{π}$

4. $\frac{2 M}{π}$

#Bar magnet

$Initial dipole moment = m = m . l (m = pole strength) arc = angle \times radius l = \frac{π}{3} \times r r = \frac{3 l}{π} Now, as the distance b / w poles = r New dipole moment = m \times r = m \times \frac{3 l}{π} = m \times l \times \frac{3}{π} = \frac{3 m}{π}$

Difficulty Level:

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23%
(2)
46%
(3)
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15%

#Earth's magnetism

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#Bar magnet

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30%
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#Magnetic field and field lines

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14%
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78%

#Earth's magnetism

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#Earth's magnetism

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#Analogy between electrostatics and magnetostatics

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#Earth's magnetism

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#Analogy between electrostatics and magnetostatics

(1)

$B = \frac{μ_{0}}{4 π} \frac{2 M}{r^{3}} B' = \frac{μ_{0}}{4 π} \frac{M}{8 r^{3}} B' = \frac{B}{16}$

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The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is
(a) $2^{- 3}$

(b) $2^{\frac{- 1}{3}}$

(d) $2^{\frac{1}{3}}$

#Bar magnet

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