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Question

# A short bar magnet placed with its axis at 30o with a uniform external magnetic field of 0.16 T experiences a torque of magnitude 0.032Nm If the bar magnet is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively

A
0.064 J,+0.064 J
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B
0.032 J,+0.032 J
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C
+0.064 J,0.128 J
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D
0.032 J,0.032 J
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Solution

## The correct option is A −0.064 J,+0.064 Jθ=30oB=0016 Tz=0.032z=mBsinθ0.032=m×12×0.16m=0.4 AmU=−→m.→BUmax=mB=0.4×0.16=0.064 TUmin=−mB=−0.4×0.16=−0.064 T

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