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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 required to keep the needle in this position will be

A

2W

B

W

C

`(W)/(sqrt(2))`

D

`sqrt(3)` W

Text Solution

Verified by Experts

The correct Answer is:
D
`W = mB cos theta = mB cos 60^(@)`
`= mB xx (1)/(2)`
`tau = mB sin theta`
`= mB sin 60^(@)`
`= sqrt(3)W`
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