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A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field vecB. The work done to rotate the loop by 30^(@) about an axis perpendicular to its plane is |
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Answer» MB Here ANGLE between `vecMandvecB` does not change and REMAIN same. `W=MBcos90^(@)` W = 0 Now if when angle between `vecBandvecM` be `theta=30^(@)`, then work done, `W=MBcos30^(@)` `thereforeW=sqrt3/2MB` |
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