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The work done to trun a magnet by 60^(...

The work done to trun a magnet by `60^(@)` from its equilibrium position is W in a uniform the torque required to hold it in that postion will be

A

w

B

`sqrt(3)/(2)` w

C

2w

D

`sqrt(3)`w

Text Solution

Verified by Experts

The correct Answer is:
D
`W=MB cos theta rarr W=(MB)/(2) rarr 2W =MB`
`tau =MB sin 60^(@)=2w.sqrt(3)/(2)=sqrt(3)W`
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