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# A bar magnet has a magnetic moment of 200Am2. The magnet is suspended in a magnetic field of 0.30 NA−1m−1. The torque required to rotate the magnet from its equilibrium position through an angle of 30o will be:30√3 Nm60 Nm60√3 Nm30 Nm

A
303 Nm
B
60 Nm
C
603 Nm
D
30 Nm
Solution
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#### Given M=200 Am2;B=0.30 NA−1M−1and θ=30oWe know that the torque→τ=→M×→B⇒|τ|=MBsinθ=200×0.3×12=100×0.3=30 Nm

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