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A bar magnet of magnetic length 0.12 m and pole strength 10 A-m is placed in a uniform magnetic field of induction 3 × 10-2 tesla. If the angle between the magnetic induction and the magnetic moment is 30°, find the magnitude of the torque acting on the magnet. |
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Answer» Data : 2l = 0.12 m, m = 10 A∙m, B = 3 × 10-2 T, θ = 30° The magnitude of the torque, τ = MB sin θ = m (2l) B sin θ = (10 A∙m) (0.12 m)(3 × 10-2 T) sin 30° = 3.6 × 10-2 × \(\cfrac12\) = 1.8 × 10-2 N∙m |
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