Practice question

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. Using the right-hand rule, if the fingers curl along the direction of the current in the loop, the thumb points perpendicular to the plane of the loop, indicating the field direction at the center. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),