JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Magnetic Properties of Matter

A bar magnet with a magnetic moment

5.0 Am^{2}

is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet form parallel to antiparallel position relative to the field direction is ______.

(A)4 J✓ Correct
(B)1 J
(C)2 J
(D)Zero

Answer: (A)

Solution

The energy of a magnetic moment in a field is

U = - M B cos θ

. In the parallel position (

θ = 0

) it is

U_{i} = - M B

, and in the antiparallel position (

θ = 18 0^{\circ}

) it is

U_{f} = + M B

. The work required equals the rise in energy,

W = U_{f} - U_{i} = 2 M B = 2 \times 5.0 \times 0.4 = 4

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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