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A magnetic needle lying parallel to a ma...

A magnetic needle lying parallel to a magnetic field requires X units of work to turn it through `60^(@)` . The torque necessary to maintain the needle in this position is

A

`sqrt(3)W`

B

W

C

`(sqrt(3))/2W`

D

`2W`

Text Solution

Verified by Experts

The correct Answer is:
A
`W=MB(cos theta_(2)-cos theta_(1))`
`=-MB(cos 60^(@)-cos 0^(@))=(MB)/2" ":. MB=2W`……..i
Torque `=MB sin 60^(@)=(2W)sin 60^(@)=(2Wxxsqrt(3))/2=sqrt(3)W`
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