JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Magnetic Properties of Matter

A bar magnet having a magnetic moment of

2.0 \times 1 0^{5} JT^{- 1}

, is placed along the direction of uniform magnetic field of magnitude

B = 14 \times 1 0^{- 5} T

. The work done in rotating the magnet slowly through

6 0^{\circ}

from the direction of field is :

(A)14 J✓ Correct
(B)8.4 J
(C)4 J
(D)1.4 J

Answer: (A)

Solution

The work done in rotating a magnetic dipole in a uniform magnetic field is

W = M B (cos θ_{1} - cos θ_{2})

. Here,

θ_{1} = 0^{\circ}

and

θ_{2} = 6 0^{\circ}

. Thus,

W = 2.0 \times 1 0^{5} \times 14 \times 1 0^{- 5} \times (cos 0^{\circ} - cos 6 0^{\circ}) = 2.0 \times 1 0^{5} \times 14 \times 1 0^{- 5} \times (1 - 0.5) = 28 \times 0.5 = 14 J

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option

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