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A magnetic needle suspended parallel to ...

A magnetic needle suspended parallel to a magnetic field requires `sqrt3J` of work to turn it through `60^@`. The torque needed to maintain the needle in this postion will be:

A

`2sqrt3N-m`

B

3 N-m

C

`sqrt3N-m`

D

`3/2N-m`

Text Solution

Verified by Experts

The correct Answer is:
B
`W=MB(costheta_(1)-costheta_(2))`
`sqrt3=MB(cos0^(@)-costheta60^(@))`
`sqrt3=MB(1-1/2)`
`MB=2sqrt3`
`tau=MBsin60^(@)`
`=2sqrt3xxsqrt3/2=3N-m`
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