1.

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….

Answer»

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

Solution :`W=mB(COS theta_1-cos theta_2)`
`=mB(COS0^@-COS30^@)`
`=mB(1-sqrt3/2)`
`=((2-sqrt3)/2)mB`


Discussion

No Comment Found

Related InterviewSolutions