1.

A short bar magnet is kept in uniform magnetic field of `0.16` T such that dipole moment vector is at angle `30^(@)` with the field . The torque acting on the bar magnet is 0.032 N m . What is the magnitude of magnetic dipole moment of the magnet ?

Answer» `tau = MB sin theta`
`0.032 = M xx 0.16 xx (1)/(2)`
`0.064 = M xx 0.16`
`M = (6.4)/(16) = 0.4 Am^(2)`


Discussion

No Comment Found

Related InterviewSolutions