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Q.

A bar magnet has a magnetic moment of 200 Am2. The magnet is suspended in a magnetic field of 0.50 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

50 N-m

b

303 N-m

c

60 N-m

d

603 N-m

answer is A.

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Detailed Solution

Torque experienced by a magnet suspended in an uniform magnetic field is given by τ=MBsinθ Given : M=200Am2; B=0.50 NA−1m−1 & θ=300 ⇒τ=200×0.50×sin30 ⇒τ=50 N−m.

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