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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

W

B

2 W

C

`sqrt(3) W`

D

`sqrt(3)/2 W`

Text Solution

Verified by Experts

The correct Answer is:
C
Work required to turn the dipole through an angle of `60^(@)` is `W=MB [cos 0^(@) - cos 60^(@)]=MB (1-1/2)`
`:. MB =2 W`
Torque required to keep the dipole in this position is
`tau =MB sin 60^(@) =(MB sqrt(3))/2=(2W xxsqrt(3))/2=sqrt(3) W`
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