Question

A magnet is parallel to a uniform magnetic field. If it is rotated by $60{}^{\circ }$ , the work done is 0.8 J. How much work is done in moving it $30{}^{\circ }$ further?

• A. $\text{0}\text{.8 }\times \text{ 1}{\text{0}}^{\text{7}}\text{erg}$
• B. 4.0 J
• C. 8 J
• D. 0.8 erg

Answer 1

Answer: (A)

$W=MB(\cos {\theta }_1-\cos {\theta }_2)$ When the magnet is rotated from $0^o$ to ${60}^o$ , then work done is $0.8J$ . $0.8=MB(\cos 0^o-\cos {60}^o)=\frac{MB}{2}$ $\Rightarrow$ $MB=1.6N-m$ In order to rotate the magnet through an angle of ${30}^o$ , $ie,$ from ${60}^o$ to ${90}^o$ , the work done is $W=MB(\cos {60}^o-\cos {90}^o)=MB\left(\frac{1}{2}-0\right)$ $=\frac{MB}{2}=\frac{1.6}{2}=0.8J=0.8\times {10}^{7}erg$