1.

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

Answer»

`SQRT3 ` W
W
`(sqrt3)/(2) W`
`2W`

SOLUTION :Work done to ROTATE a magnetic needle from `theta_1 = 0^@ ` to `theta_2 = 60^@`
W= mB`[cos 0^@ - cos 60^@ ] = -(mB)/(2)`
Torque required to MAINTAIN magnetic needle at `theta = 60^@ ` will be `tau = mB sin 60^@ = (mBsqrt3)/(2)`
Hence `|tau|= sqrt3 |W|`


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