A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60°$ . The torque required to maintain the needle in this position will be
AIEEE 2003, Diffcult
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(a)$W = MB(\cos {\theta _1} - \cos {\theta _2}) = MB(\cos {0^o} - \cos {60^o})$
$ = MB\,\left( {1 - \frac{1}{2}} \right) = \frac{{MB}}{2}$
and $\tau = MB\sin \theta = MB\sin {60^o} = MB\frac{{\sqrt 3 }}{2}$
$\therefore \;\tau = \left( {\frac{{MB}}{2}} \right)\sqrt 3 \Rightarrow \tau = \sqrt 3 \,W$
$ = MB\,\left( {1 - \frac{1}{2}} \right) = \frac{{MB}}{2}$
and $\tau = MB\sin \theta = MB\sin {60^o} = MB\frac{{\sqrt 3 }}{2}$
$\therefore \;\tau = \left( {\frac{{MB}}{2}} \right)\sqrt 3 \Rightarrow \tau = \sqrt 3 \,W$
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