1.

A bar magnet of moment 10 A.m2 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 39 μT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 10°.

Answer»

Data : μ = 10 A.m2 , Bh = 3.9 × 10-5 T, θ = 10° 

The magnitude of the torque is τ = – μBh sin θ 

= (10)(3.9 × 10-5) sin 10° 

= (3.9 × 10-4)(0.1736) = 6.770 × 10-5 N.m



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