The magnetic moment of a bar magnet is 0.5 Am². It is suspended in a uniform magnetic field of 8 × 10^–2 T. The work done in rotating it from its most stable to most unstable position is :

The correct answer is 8 × 10^–2 J.

We are given the following information:

Magnetic moment, $m$ = 0.5 Am²
Uniform magnetic field, $B$ = 8 × 10^–2 T

Step 1: Identify the most stable and most unstable positions.

The potential energy of a bar magnet in a uniform magnetic field is given by:

$U=-mB\cos \theta$

where $\theta$ is the angle between the magnetic moment vector and the magnetic field direction.

• Most stable position: When $\theta =\mathrm{0°}$ (magnet aligned parallel to the field), the potential energy is minimum:

$U_1=-mB\cos \mathrm{0°}=-mB$

• Most unstable position: When $\theta =\mathrm{180°}$ (magnet aligned anti-parallel to the field), the potential energy is maximum:

$U_2=-mB\cos \mathrm{180°}=+mB$

Step 2: Calculate the work done.

The work done in rotating the magnet from the most stable (θ = 0°) to the most unstable (θ = 180°) position equals the change in potential energy:

$W=U_2-U_1=mB-(-mB)=2mB$

Step 3: Substitute the values.

$W=2\times 0.5\times 8\times {10}^{-2}$

$W=1\times 8\times {10}^{-2}$

$W=8\times {10}^{-2}\text{J}$

Therefore, the work done in rotating the bar magnet from its most stable to its most unstable position is 8 × 10^–2 J.