Question

0.8 J work is done in rotating a magnet by ${60}^o$, placed parallel to a uniform magnetic field. How much work is done in rotating it ${30}^o$ further?

• A. $0.8\times {10}^{7}erg$
• B. $0.8erg$
• C. $8J$
• D. $0.4J$

Answer 1

The correct option is A $0.8\times {10}^{7}erg$

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