1.

A bar magnet having a magnetic moment of 2.0 xx 10^(-4) J//T is free to rotate in a horizontal plane. A horizontal magnetic field B = 5 xx 10^(-5) T exists in the space. Then the work done in rotating the magnet slowls from a direction parallel to the field to a direction 60^(@) from the field is

Answer»

0.1 J
0.3 J
0.5 J
0.7 J

Solution :The work done by the external agent = change in POTENTIAL energy
`= (-MB costheta_(2)) - (-MBcostheta_(1))`
`=-M(cos60^(@)-cos0^(@))`
`=1/2MB = 1/2xx(2.0xx10^(4)J//T)(5xx10^(-5)T = 0.5J`


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