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A circular current loop of magnetic moment M is in an arbitary orientation is an external magnetic field vec B. The work done to rotate the loop by 30^@ about an axis perpendicular to its plane is…. |
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Answer» MB `=mB(COS0^@-COS30^@)` `=mB(1-sqrt3/2)` `=((2-sqrt3)/2)mB` |
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