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Find the magnitude of the magnetic moment of a magnet if a couple exerting torque 0.5 N.m is required to hold the magnet with its axis perpendicular to a uniform magnetic field of induction 2 × 10-3T |
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Answer» Data : τ = 0.5 N∙m, B = 2 × 10-3 T, θ = 90° τ = MB sin θ ∴ M = \(\cfrac{\tau}{B\,sin\theta}\) = \(\cfrac{0.5\,N.m}{(2\times10^{-3}T)sin \,90^\circ}\) = 250 A∙m2 The magnitude of the magnetic moment of the magnet is 250 A∙m2 . |
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