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InterviewSolution
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A bar magnet of moment 10 A∙m is suspended such that it can rotate freely in a horizontal plane. The horizontal component of the Earth’s magnetic field at the place is 36 µT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 8C and magnetic potential energy. |
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Answer» Data : M = 10 A∙m2, Bh = 36µT = 3.6 × 10-5 T, θ = 8° The magnitude of the torque is τ = MBh sin θ = (10) (3.6 × 10-5) sin 8° = (36 × 10-5) (0.1391) = 5.007 × 10-5 N∙m The magnetic potential energy of the bar magnet is Uθ = MBh cos θ = (36 × 10-5) cos 8° = (36 × 10-5) (0.99) = 3.564 × 10-4 J |
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