Physics - Magnetism Question with Solution | TestHub

Work done in rotating a dipole

$\mathrm{W}=\mathrm{M}\mathrm{B}\mathrm{ }\left(\cos {\mathrm{\theta }}_1-\cos {\mathrm{\theta }}_2\right)$

To rotate it from$90°$to$0°$

${\mathrm{\theta }}_1=90°$${\mathrm{\theta }}_2=0°$

$\mathrm{W}=\mathrm{M}\mathrm{B}\left(\cos 90°-\cos 0°\right)$

${\mathrm{W}}_1=\mathrm{M}\mathrm{B}\left(0-1\right)=-\mathrm{M}\mathrm{B}$

${\mathrm{W}}_1=-\mathrm{M}\mathrm{B}$.........(i)

To rotate it from$0°$to$-90°$in same manner.

${\mathrm{\theta }}_1=0°$${\mathrm{\theta }}_2=-90°$

${\mathrm{W}}_2=\mathrm{M}\mathrm{B}\left(\cos 0°-\cos \left(-90°\right)\right)$

${\mathrm{W}}_2=\mathrm{M}\mathrm{B}$........(ii)

Total workdone

$\mathrm{W}={\mathrm{W}}_1+{\mathrm{W}}_2$

$\mathrm{W}=-\mathrm{M}\mathrm{B}+\mathrm{M}\mathrm{B}$

$\mathrm{W}=0$