1.

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

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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