Two parallel wires \( 0.12 \, \text{m} \) apart carry currents of \( 4 \, \text{A} \) and \( 7 \, \text{A} \) in the sam
**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 4 × 7/2 π × 0.12) = (112 × 10⁻⁷/0.24) = 4.67 × 10⁻⁶ N/m . Using F
Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer