JEE MAIN - Physics (2003 - No. 16)
A magnetic needle lying parallel to a magnetic field requires $$W$$ units of work to turn it through $${60^ \circ }.$$ The torque needed to maintain the needle in this position will be :
$$\sqrt 3 \,W$$
$$W$$
$${{\sqrt 3 } \over 2}W$$
$$2W$$
Explanation
$$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$$
$$ = MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)$$
$$ = MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}$$
$$\therefore$$ $$\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }$$
$$ = \sqrt 3 {{MB} \over 2} = \sqrt 3 W$$
$$ = MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)$$
$$ = MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}$$
$$\therefore$$ $$\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }$$
$$ = \sqrt 3 {{MB} \over 2} = \sqrt 3 W$$
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