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  4. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be

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Q. A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^°$. The torque needed to maintain the needle in this position will be

AIEEEAIEEE 2003Magnetism and Matter

Solution:

$W = −MB(cos\, θ_2 − cos \,θ_1)$
Initially magnetic needle is parallel to a magnet field, therefore
$θ_1 = 0,$
$θ_2 = 60^°$
$∴ W = −MB(cos\, 60^° − cos \,0^°)$
$=MB$
$e =MB\, sin 60^° =ZW\times\sqrt{3}/ 2=\sqrt{3}W$