Consider a planar current loop of turns

PHYSICS · Multi-part · CBSE · Physics · Moving Charges and Magnetism · Moving Charges and Magnetism

Problem

Consider a planar current loop of NN turns, area AA, carrying current II, placed in a uniform magnetic field BB. The loop can rotate about an axis through its centre and within its own plane. (i) Starting from the expression for the torque acting on the loop, derive an expression for the potential energy of the magnetic dipole as a function of the angle θ\theta between its magnetic moment and the field direction. (ii) Using the result of part (i), obtain a general formula for the work done in rotating the dipole from an initial angle θ1\theta_1 to a final angle θ2\theta_2. (iii) For N=50N = 50, A=8.0×10−4 m2A = 8.0\times10^{-4}\,\text{m}^2, I=0.30 AI = 0.30\,\text{A} and B=0.15 TB = 0.15\,\text{T}, calculate the work required to rotate the loop from θ1=90∘\theta_1 = 90^{\circ} to θ2=30∘\theta_2 = 30^{\circ}.

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