Question

A magnet of moment $4.0$ ${\text{Am}}^2$ is held in a uniform magnetic field $5.0\times {10}^{-4}$ Tesla with the magnetic moment vector making an angle of ${30}^{\circ }$ with the field. The work done in increasing the angle from ${30}^{\circ }\text{to}{45}^{\circ }$ is nearly

• A. $3.18\times {10}^{-4}\text{J}$
• B. $4.3\times {10}^{-4}\text{J}$
• C. $7.3\times {10}^{-4}\text{J}$
• D. Zero

Answer 1

The correct option is A $3.18\times {10}^{-4}\text{J}$

Work done as angle changes from ${\theta }_1$ to ${\theta }_2$ is
$W=-MB(\cos {\theta }_2-\cos {\theta }_1)$
$W=4\times 5\times {10}^{-4}[\cos {30}^{\circ }-\cos {45}^{\circ }]$
$W=20\times {10}^{-4}[\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}]$
$W=3.18\times {10}^{-4}\text{J}$