A rectangular loop with length $0.12,\text{m}$, width $0.08,\text{m}$ carries a current of $4,\text{A}$. It is placed in a uniform magnetic field of magnitude $0.35,\text{T}$ directed along the $+x$‑axis, and the loop’s normal makes an angle of $30^{\circ }$ with the field. What is the magnitude of the torque acting on the loop?

Step-by-step solution

1. Compute the area of the loop: $A=(0.12,\text{m})(0.08,\text{m})=9.6\times 10^{-3},{\text{m}}^2$.
2. Find $sin\theta$ for $\theta =30^{\circ }$: $sin30^{\circ }=0.5$.
3. Use the torque formula for a single‑turn loop $\tau =IABsin\theta$ to obtain $\tau =(4,\text{A})(9.6\times 10^{-3},{\text{m}}^2)(0.35,\text{T})(0.5)$.
4. Calculate: $4\times 9.6\times 10^{-3}=3.84\times 10^{-2}$, $3.84\times 10^{-2}\times 0.35=1.344\times 10^{-2}$, $1.344\times 10^{-2}\times 0.5=6.72\times 10^{-3},\text{N⋅m}$.
5. State the final result $\tau =6.7\times 10^{-3},\text{N⋅m}$.

Concept tested: Torque on a current loop

Published 21 September 2026 · How this solution was prepared and checked