A rectangular loop of width $8\text{cm}$ and height $12\text{cm}$ carries a current of $3.0\text{A}$. The loop is placed in a uniform magnetic field of magnitude $0.60\text{T}$ that makes an angle of $45^{\circ }$ with the normal to the plane of the loop. Calculate the torque on the loop and state the direction in which the torque tends to rotate the loop.

Step-by-step solution

1. Calculate the area of the loop: $A=(0.08\text{m})(0.12\text{m})=9.6\times 10^{-3}{\text{m}}^2$.
2. Use the torque formula for a current loop: $\tau =NIAB\sin \theta$, with $N=1$.
3. Substitute the values: $\tau =1\times 3.0\times 9.6\times 10^{-3}\times 0.60\times \sin 45^{\circ }$.
4. Evaluate $\sin 45^{\circ }=\frac{\sqrt{2}}{2}=0.707$.
5. Compute: $3.0\times 9.6\times 10^{-3}=2.88\times 10^{-2}$; multiply by $0.60$ gives $1.728\times 10^{-2}$; multiply by $0.707$ gives $1.22\times 10^{-2}\text{N⋅m}$.
6. The torque vector is given by $\vec{\tau }=\vec{\mu }\times \vec{B}$, so it acts to rotate the loop so that its normal aligns with $\vec{B}$.
7. Thus the loop tends to rotate in the direction that reduces the angle between the normal and the magnetic field.

Concept tested: Torque on a current loop in a magnetic field

Published 17 September 2026 · How this solution was prepared and checked