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. tau = N I A B sin θ , where θ = 60° to the plane means sin 30° with the normal. tau = 10 × 3 × 0.02 × 0.5 × sin 60° = 0.3 × 0.866 = 0.2598 ≈ 0.26 N m . Using F = q v B sinθ,