The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^{2}$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is :

(B) $8 \times 10^{-2} \mathrm{~J}$

(A) $16 \times 10^{-2} \mathrm{~J}$

(C) $4 \times 10^{-2} \mathrm{~J}$

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Magnetic Properties of Matter

The magnetic moment of a bar magnet is

0.5 Am^{2}

. It is suspended in a uniform magnetic field of

8 \times 1 0^{- 2} T

. The work done in rotating it from its most stable to most unstable position is :

(A) $16 \times 1 0^{- 2} J$
(B) $8 \times 1 0^{- 2} J$ ✓ Correct
(C) $4 \times 1 0^{- 2} J$
(D)Zero

Answer: (B)

Solution

Potential energy of a magnetic dipole in a field:

U = - m B cos θ

.
Most stable position (

θ = 0^{\circ}

U_{i} = - m B

.
Most unstable position (

θ = 18 0^{\circ}

U_{f} = + m B

.
Work done

= U_{f} - U_{i} = 2 m B = 2 (0.5) (8 \times 1 0^{- 2}) = 8 \times 1 0^{- 2} J

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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