A rectangular current loop of length $0.12,\text{m}$ and width $0.05,\text{m}$ carries a current of $5,\text{A}$. It is placed in a uniform magnetic field of magnitude $0.8,\text{T}$. The normal to the plane of the loop makes an angle of $30^{\circ }$ with the direction of the magnetic field. Find the magnitude of the torque acting on the loop and specify its direction.

Step-by-step solution

1. Calculate the area of the loop: $A=l\times w=0.12,\text{m}\times 0.05,\text{m}=0.006,{\text{m}}^2$.
2. Compute $sin\theta$ for $\theta =30^{\circ }$: $sin30^{\circ }=0.5$.
3. Use the torque formula for a single-turn loop: $\tau =IABsin\theta$ (with $N=1$).
4. Substitute $I=5,\text{A}$, $A=0.006,{\text{m}}^2$, $B=0.8,\text{T}$ and $sin\theta =0.5$.
5. Compute $IA=5\times 0.006=0.03$.
6. Multiply by $B$: $0.03\times 0.8=0.024$.
7. Multiply by $sin\theta$: $0.024\times 0.5=0.012,\text{N}\cdot \text{m}$.
8. Determine direction: the torque tends to rotate the loop so that its normal aligns with the magnetic field; for a clockwise current when viewed from above, this rotation is clockwise.
9. State the result as $0.012,\text{N}\cdot \text{m}$ clockwise.

Concept tested: torque on current loop in magnetic field

Published 15 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked