A solenoid has 500 turns, length 0.20 m, radius 0.05 m, and carries a current of 3.0 A. It is placed in a uniform magnetic field of 0.10 T that is perpendicular to the axis of the solenoid.

Step-by-step solution

1. Compute the cross‑sectional area of the solenoid: $A=\pi a^2=\pi (0.05{)}^2=0.00785398;{\text{m}}^2$.
2. Use the formula for magnetic moment of a solenoid: $\mu =NIA$ with $N=500$, $I=3.0;\text{A}$, $A=0.00785398;{\text{m}}^2$.
3. Calculate $\mu =500\times 3.0\times 0.00785398=11.78097;{\text{A⋅m}}^2$, round to $11.8;{\text{A⋅m}}^2$.
4. Since the external field is perpendicular to the solenoid axis, the angle $\theta$ between $\mu$ and $B$ is $90^{\circ }$.
5. Use the torque formula $\tau =\mu Bsin\theta$. With $sin90^{\circ }=1$, $\tau =11.78\times 0.10=1.178;\text{N⋅m}$, round to $1.18;\text{N⋅m}$.
6. The potential energy of a magnetic dipole is $U=-\mu Bcos\theta$. With $cos90^{\circ }=0$, $U=0;\text{J}$.
7. The torque direction follows the right–hand rule for $\mu \times B$ and will rotate the solenoid so that its axis becomes parallel to the external field.
8. Thus $\mu =11.8;{\text{A⋅m}}^2$, $\tau =1.18;\text{N⋅m}$, $U=0;\text{J}$, and the torque tends to align the solenoid with the field.

Concept tested: Torque on magnetic dipole in external field

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked