Question

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

• A. \( 16 \times 10^{-2} \, \text{J} \)
• B. \( 8 \times 10^{-2} \, \text{J} \)
• C. \( 4 \times 10^{-2} \, \text{J} \)
• D. \( \text{Zero} \)

Answer 1

The Correct Option is B

Magnetic Potential Energy:
The potential energy U of a magnetic moment m in a uniform magnetic field B is given by:
U = -mB cos θ.

Stable and Unstable Equilibrium Positions:
1. At stable equilibrium (θ = 0°):
U_stable = -mB cos 0° = -mB.

2. At unstable equilibrium (θ = 180°):
U_unstable = -mB cos 180° = +mB.

Work Done:
The work done in rotating the magnet from the stable to the unstable position is the change in potential energy:
W = ΔU = U_unstable - U_stable = mB - (-mB) = 2mB.

Substitute Values:
Given:
m = 0.5 Am², B = 8 × 10^-2 T,
W = 2 × 0.5 × 8 × 10^-2 = 8 × 10^-2 J.

Answer: 8 × 10^-2 J