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A short bar magnet placed with its axis at 30^(@) with a uniform external magnetic field of 0.25T. T experiences a torque of magnitude equal to 4.5 xx 10^(-2)J. What is the magnitude of magntic moment of the magnet? |
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Answer» Solution :When magnetic dipole moment `overset(to) (m)` of a bar magnet makes angle `theta` with external uniform magnetic field `overset(to) (B)` TORQUE `overset(to) (TAU)` exerted on it is GIVEN by, `overset(to)(tau ) = overset(to) xx overset(to)(B)""...(1)` `therefore tau = mB SIN theta""...(2)` `therefore m= (tau)/( B sin theta)` `= (4.5 xx 10^(-2) )/( (0.25) (sin 30^@) )` `= (0.045)/( (0.25) (0.5) )` `therefore m= 0.36` (ampere)(metre) `""^(2)` |
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