Question

A short magnet placed with its axis at ${30}^{o}C$ with external magnetic field 800 G, experience a torque of 0.016 Nm. Work done in moving it from most stable to unstable position is:

• A. 0.032 J
• B. 0.064 J
• C. 0.128 J
• D. 0.4 J

Answer 1

The correct option is B 0.064 J

Given,

$\theta ={30}^0$

$\tau =0.016Nm$

$B=800G=800\times {10}^{-4}T$

Torque acting on a magnet,

$\tau =MBsin\theta$

$M=\frac{\tau }{Bsin\theta }=\frac{0.016}{800\times {10}^{-4}\times sin{30}^0}$

$M=0.40A{m}^2$

Work done by moving the magnet from most stable to unstable position,

$W=-MB(cos{\theta }_2-cos{\theta }_1)$

$W=-MB(cos{180}^0-cos{0}^0)=-MB(-1-1)$

$W=2MB=2\times 0.40\times 800\times {10}^{-4}$

$W=0.064J$

The correct option is B.