A bar magnet has a magnetic moment of 100 Am2. The magnet is suspended in a magnetic field of 0.60 NA-1m-1. The torque required to rotate the magnet from its equilibrium position through an angle of 300 will be:
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a
30 N-m
b
303N-m
c
60 N-m
d
603N-m
answer is A.
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Detailed Solution
Torque required by a magnet suspended is an uniform magnetic field is given by τ=MB sinθM=100 Am2 ; B=0.6 NA-1m-1 & θ=300τ=(100)(0.6)sin300 τ=30 N-m .