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A magnetic needle lying parallel to a ma...

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

A

`sqrt3W`

B

W

C

`(sqrt3)/(2)W`

D

2W

Text Solution

Verified by Experts

The correct Answer is:
A
`W= Delta U= - MB (cos theta_(f)- cos theta_(i))`
`= - MB (cos 60^(@)- cos 0^(@))`
`W= (MB)/(2)`
`tau = MB sin theta rArr tau = MB (sqrt3)/(2)`
`therefore tau= sqrt3 W`
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