A rectangular loop of width 10 cm and height 20 cm carries a current of 4 A. It is placed in a uniform magnetic field of 0.5 T that makes an angle of 30° with the normal to the loop. Find (i) the torque on the loop, (ii) the potential energy, and state the direction of the torque with respect to the field.

Step-by-step solution

1. Compute the area of the loop: $A=0.10,m\times 0.20,m=0.02,m^2$.
2. Find the magnetic dipole moment: $\mu =IA=4\times 0.02=0.08,A,m^2$.
3. Torque magnitude: $\tau =\mu Bsin\theta =0.08\times 0.5\times \sin 30^{\circ }=0.08\times 0.5\times 0.5=0.02,Ncdotm$.
4. Potential energy: $U=-\mu Bcos\theta =-0.08\times 0.5\times \cos 30^{\circ }=-0.08\times 0.5\times 0.866\approx -0.0346,Japprox-3.5\times 10^{-2},J$.
5. Direction: the torque acts to rotate the loop so that the dipole moment vector $mu$ aligns with the magnetic field $\mathbf{B}$ (i.e., it tends to rotate the loop to minimize the angle between $mu$ and $B$).

Concept tested: torque and potential energy of a current loop

Published 15 September 2026 · How this solution was prepared and checked