A rectangular current loop has length $0.09,\text{m}$, width $0.04,\text{m}$ and carries a current of $6.0,\text{A}$. It is placed in a uniform magnetic field of magnitude $0.50,\text{T}$ making an angle of $60^{\circ }$ between its normal and the field direction. Find the magnitude of the torque acting on the loop.

Step-by-step solution

1. Find area $A=l,w=0.09\times 0.04=3.6\times 10^{-3},{\text{m}}^2$
2. Use torque formula $\tau =IABsinphi$ with $\varphi =60^{\circ }$
3. Compute $sin60^{circ}=0.8660$
4. Multiply: $IAB=6.0\times 3.6\times 10^{-3}\times 0.50=0.0108,\text{N⋅m}$
5. Apply sine factor: $\tau =0.0108\times 0.8660=9.35\times 10^{-3},\text{N⋅m}$

Concept tested: Torque on current loop in magnetic field

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked