Solveeit Logo
Login

Question

Physics Question on The Bar Magnet

The magnetic moment of a bar magnet is 0.5 Am20.5 \, \text{Am}^2. It is suspended in a uniform magnetic field of 8×10−2 T8 \times 10^{-2} \, \text{T}. The work done in rotating it from its most stable to most unstable position is:

A

16×10−2 J16 \times 10^{-2} \, \text{J}

B

8×10−2 J8 \times 10^{-2} \, \text{J}

C

4×10−2 J4 \times 10^{-2} \, \text{J}

D

Zero\text{Zero}

Answer

8×10−2 J8 \times 10^{-2} \, \text{J}

Explanation

Solution

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 - Ustable = mB - (-mB) = 2mB.

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

Answer: 8 × 10 -2 J