A "bar" magnet of magnetic moment 1.5J//...

A "bar" magnet of magnetic moment `1.5J//T` is aligned with the direction of a uniform magnetic field of `0.22 T`. The work done in turning the magnet so as to align its magnetic moment opposite to the field and the torque acting on it in this position are respectively.

`0.33J,0.33N-m`

`0.66J,06.66B-m`

`0.33J,0`

`0.66J,0`

Text Solution

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The correct Answer is:

`tau=MB sin theta , W=MB(costheta_(1)-costheta_(2))`

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NARAYNA-MAGNETISM-LEVEL-I(C.W)

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