Question

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 36 µT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 8° and magnetic potential energy.

Answer 1

Data : M = 10 A∙m^2, 

B_h = 36 µT = 3.6 × 10^-5 T, 

θ = 8°

The magnitude of the torque is

τ = MB_h sin θ

=(10)(3.6 × 10^-5)sin8°

=(36 × 10^-5)(0.1391) = 5.007 × 10^-5 N∙m

The magnetic potential energy of the bar magnet is

U_θ = MB_h cos θ 

=(36 × 10^-5)cos8° 

=(36 × 10^-5)(0.99) = 3.564 × 10^-4 J