A rectangular current loop of width $4.0\text{cm}$ and height $6.0\text{cm}$ carries a current of $2.0\text{A}$. The loop is placed in a uniform magnetic field of magnitude $0.50\text{T}$ that makes an angle of $60^{\circ }$ with the normal to the plane of the loop. Calculate (i) the torque on the loop and (ii) the magnetic potential energy. State the direction of the torque relative to the field.

Step-by-step solution

1. Step 1: Compute the area $A$ of the loop: $A=lw=0.06\text{m}\times 0.04\text{m}=2.4\times 10^{-3}{\text{m}}^2$.
2. Step 2: Calculate the magnetic dipole moment $\mu =IA=2.0\times 2.4\times 10^{-3}=4.8\times 10^{-3}{\text{A,m}}^2$.
3. Step 3: Torque magnitude is $\tau =\mu Bsin\theta$, where $\theta =60^{\circ }$. Plug in values: $\tau =4.8\times 10^{-3}\times 0.50\times \sin 60^{\circ }$.
4. Step 4: Evaluate $sin60^{\circ }=\sqrt{3}/2\approx 0.866$ and compute $\tau =4.8\times 10^{-3}\times 0.50\times 0.866=1.73\times 10^{-3}\text{N,m}$.
5. Step 5: Potential energy is $U=-\mu Bcos\theta$. Compute $cos60^{\circ }=0.5$ and thus $U=-4.8\times 10^{-3}\times 0.50\times 0.5=-1.20\times 10^{-3},\text{J}$? Wait calculation: $4.8\times 10^{-3}\times 0.50=2.4\times 10^{-3}$, times $0.5$ gives $1.20\times 10^{-3}$, with minus sign gives $-1.20\times 10^{-3},\text{J}$. However the problem requires energy in joules; re-evaluate: $U=-\mu Bcos\theta =-4.8\times 10^{-3}\times 0.50\times 0.5=-1.20\times 10^{-3},\text{J}$. This is correct.
6. Step 6: State direction: the torque tends to rotate the loop so that its magnetic moment aligns with the external field, i.e., reduces the angle between $mu$ and $B$.

Concept tested: Torque on magnetic dipole

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