A rectangular current loop of width $10,\text{cm}$ and height $15,\text{cm}$ carries a current of $4,\text{A}$. It is placed in a uniform magnetic field of magnitude $0.50,\text{T}$ that makes an angle of $30^{\circ }$ with the normal to the plane of the loop. (i) Calculate the torque on the loop. (ii) Calculate the magnetic potential energy of the loop. State the direction of the torque relative to the field.

Step-by-step solution

1. Compute the area of the loop: $A=wcdoth=0.10,\text{m}\times 0.15,\text{m}=0.015,{\text{m}}^2$.
2. Determine the magnetic dipole moment: $\mu =I,A=4,\text{A}\times 0.015,{\text{m}}^2=0.06,{\text{A,m}}^2$.
3. Use the torque formula for a magnetic dipole: $\tau =\mu Bsin\theta$. Plugging in: $\tau =0.06\times 0.50\times \sin 30^{\circ }=0.015,\text{N,m}$.
4. Use the potential energy formula: $U=-\mu Bcos\theta$. Thus $U=-0.06\times 0.50\times \cos 30^{\circ }\approx -2.6\times 10^{-2},\text{J}$.
5. The torque acts to reduce $\theta$, i.e., it rotates the loop so that its normal aligns with the field.

Concept tested: Torque on current loop

Published 19 September 2026 · How this solution was prepared and checked