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The work done in rotating a magnet of po...

The work done in rotating a magnet of pole strength `1 A-m` and length `1 cm` through an angle of `60^(@)` from the magnetic meridian is `(H=30A//m)`

A

`9.42xx10^(-8)J`

B

`3.14xx10^(-8)J`

C

`18.84xx10^(-8)J`

D

`10xx10^(-8)J`

Text Solution

Verified by Experts

The correct Answer is:
C
`W=mxx2l B[cos theta_(1)-cos theta_(2)]`
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