A coil having 10 A m² magnetic moment is placed in a vertical plane and is free to rotate about its horizontal axis which coincides with its diameter. A uniform magnetic field of 2 T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The co…

Question

A coil having 10 A m² magnetic moment is placed in a vertical plane and is free to rotate about its horizontal axis which coincides with its diameter. A uniform magnetic field of 2 T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field. The moment of inertia of the coil is 0.1 kg m². What will be its angular speed?

Accepted Answer

(1) 20 rad s⁻¹. Given: m=10 A m², B=2 T, I=0.1 kg m², and a quarter turn between the aligned position and the perpendicular one. Idea: a magnetic dipole in a field has energy U=-mB cos θ, so between θ=0 and θ=90° the coil exchanges |Δ U|=mB with the field. |Δ U|=mB=10×2=20 J. All of it goes into rotational kinetic energy: 1/2Iω²=20. ω²=(2×20)/(0.1)=400, so ω=20 rad s⁻¹. Worth noticing: as the stem is printed the coil starts with its axis along the field, which is the position of lowest energy and zero torque, so this 20 J has to be supplied rather than released — the field gives it up when the coil swings the other way, into alignment. The size of the exchange, and so the answer, is the same either way.

Step-by-step solution

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