A rectangular current loop of length $0.15,m$ and width $0.08,m$ carries a current of $4,A$ counter‑clockwise when viewed from above. It is placed in a uniform magnetic field of $0.3,T$ making an angle of $25^{c}irc$ between the normal to the loop and the field direction. What is the magnitude of the torque on the loop and in which direction does it act?

Step-by-step solution

1. The magnetic torque on a current loop is $\tau =NIAB\sin \theta$, where $N$ is the number of turns (1 here), $A$ is the area of the loop, and $\theta$ is the angle between the normal vector and the magnetic field.
2. Compute the area: $A=l\times w=0.15m\times 0.08m=0.012m^2$.
3. Substitute values: $\tau =(1)(4A)(0.012m^2)(0.3,T)\sin 25^{c}irc$.
4. Evaluate $\sin 25^{c}irc\approx 0.4226$.
5. Then $\tau =4\times 0.012\times 0.3\times 0.4226=0.006086N\cdot m\approx 6.1\times 10^{-3}N\cdot m$.
6. Direction: The torque vector is given by $\mathbf{N}\times \mathbf{B}$ (right‑hand rule). Because the current is counter‑clockwise as seen from above, the torque tends to rotate the loop clockwise about its centre to align the normal with $\mathbf{B}$.
7. Thus the torque magnitude is $6.1\times 10^{-3}N\cdot m$ and acts clockwise about the centre of the loop.

Concept tested: Torque on a current loop

Published 18 September 2026 · How this solution was prepared and checked