1.

A short bar magnet placed with its axis at 30^@ with a uniform magnetic field of 0.16 T experiences a torque of magnitude 0.032 J. The magnetic dipole moment of the bar magnet is

Answer»

`0.23 J T^(-1)`
`0.40 J T^(-1)`
`0.80 J T^(-1)`
0

Solution :`VEC m = (tau)/(B SIN theta) = (0.032)/(0.16xxsin 30^@) = 0.40 J T^(-1)`


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