Question

A small bar magnet placed with its axis at $30°$ with an external field of $0.06T$experiences a torque of $0.018Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is:

• A. $7.2\times {10}^{-2}J$
• B. $6.4\times {10}^{-2}J$
• C. $9.2\times {10}^{-3}J$
• D. $11.7\times {10}^{-3}J$

Answer 1

The correct option is A

$7.2\times {10}^{-2}J$

Step 1. Given data:

Angle made with horizontal axis $\theta =30°$

Field applied, $B=0.06T$

Torque experienced,$\tau =0.018Nm$

Step 2. Finding work done:

Torque is given by, $\tau =MB\sin \theta$

$\Rightarrow 0.018=M\times 0.06\times \sin 30°$

$\Rightarrow M=\frac{0.018}{0.06\times \sin 30°}=0.6A{m}^2$

The bar is in stable position when $\theta =0°$ and in unstable position when $\theta =180°$

Work done to rotate it from its stable to unstable equilibrium position is $W=-MB\cos 180°-(-MB\cos 0°)$

$\Rightarrow W=2MB=2\times 0.6\times 0.06$

$\Rightarrow W==0.072J=7.2\times {10}^{-2}J$

Hence, option A is correct.