Question

A magnet is suspended at an angle ${60}^{\circ }$ in an uniform external magnetic field of $5\times {10}^{-4}\text{T}$. The work done in bringing the magnet in the direction of the magnetic field is (The magnetic moment $=20{\text{Am}}^2$)

• A. $-5\times {10}^{-3}\text{T}$
• B. $+5\times {10}^{-3}\text{T}$
• C. $-3\times {10}^{-3}\text{T}$
• D. $+3\times {10}^{-3}\text{T}$

Answer 1

The correct option is A $-5\times {10}^{-3}\text{T}$

Workdone by the magnetic field , $W=MB[\cos {\theta }_1-\cos {\theta }_2]$

Here, ${\theta }_1={60}^{\circ };{\theta }_2=0^{\circ }$

Therefore, $W=20\times 5\times {10}^{-4}\times [\cos {60}^{\circ }-\cos 0^{\circ }]$

$W={10}^{-2}\times [\frac{1}{2}-1]=-5\times {10}^{-3}\text{J}$