1.

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

Answer»

MB
`sqrt3/2MB`
`(MB)/2`
zero

Solution :
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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