Torque on a current loop

Chop the loop into elements dL and use the previous topic's result, dF = I dL × B. Because B is identical at every element, it comes outside the sum: F = I (∮ dL) × B. But ∮ dL is the vector sum of the little steps that walk right round a closed path and return to the start, so it is exactly zero, and so is F. Take a rectangle 6.0 cm by 4.0 cm carrying 3.0 A in B = 0.20 T to watch it happen: the two 6.0 cm sides carry current perpendicular to B and each feels I a B = 3.0 × 0.060 × 0.20 = 0.036 N, in opposite directions; the two 4.0 cm sides feel forces that are also equal and opposite. Everything cancels. What does not cancel is where those forces act — the 0.036 N pair is applied on opposite sides of the pivot, and a pair like that is a couple.

Pivot that rectangle about a vertical axis through the midpoints of its two horizontal sides, with B horizontal, and let θ be the angle between the loop's normal and B. The two vertical sides of length a = 6.0 cm always run perpendicular to B, so each keeps its force I a B = 0.036 N whatever θ is, and those forces stay horizontal and perpendicular to B. Each acts a distance (b/2) sin θ = 0.020 sin θ m from the axis, so the two torques add rather than cancel: τ = 2 × (I a B)(b/2) sin θ = I (ab) B sin θ = I A B sin θ. With A = 2.4 × 10⁻³ m² that is 1.44 × 10⁻³ sin θ N m — 1.44 × 10⁻³ N m at θ = 90°, 7.2 × 10⁻⁴ N m at θ = 30°, zero at θ = 0. The forces on the horizontal sides point along the pivot axis, so they exert no torque about it.

Notice what dropped out. The sides a and b entered only as their product, the area. Stack N identical turns and each contributes the same couple, giving N I A B sin θ. So define μ = N I A n̂, where n̂ is the loop's normal, pointed by curling the right hand along the current so the thumb gives n̂. Then τ = μ × B, and the cross product carries the sin θ and the rotation axis together. The area is all that matters, not the shape: a circular coil of the same 2.4 × 10⁻³ m² carrying the same current, which needs radius r = √(A/π) = 2.76 × 10⁻² m, feels exactly the same torque in the same field. Units check: A m² × T = A m² × N A⁻¹ m⁻¹ = N m. The same μ describes a galvanometer coil, a motor armature, a compass needle and an orbiting electron.

The torque always turns μ towards B, so turning it away costs work, and that work is stored: integrating μB sin θ through dθ gives U = −μ⋅B = −μB cos θ. It is least, −μB, at θ = 0, and greatest, +μB, at θ = 180°, so θ = 0 is stable and θ = 180° is unstable — nudge the reversed loop and it flips right over. For μ = 0.080 A m² in B = 0.30 T, μB = 0.024 N m, so flipping the loop end for end costs 2μB = 0.048 J. Near alignment cos θ ≈ 1 − θ²/2 and the restoring torque is τ ≈ −μBθ, the signature of simple harmonic motion: with a moment of inertia Iᵣₒₜ = 1.5 × 10⁻⁵ kg m², ω = √[(0.024 N m) ÷ (1.5 × 10⁻⁵ kg m²)] = 40 rad s⁻¹, so the loop oscillates at 6.4 Hz about θ = 0 rather than parking there.

A motor is this torque harnessed, and the sin θ is the problem. Take an armature of N = 200 turns, A = 2.0 × 10⁻³ m², carrying 4.0 A in B = 0.35 T: μ = 1.6 A m² and the peak torque is μB = 0.56 N m. Averaged over half a turn the mean of sin θ is 2/π = 0.637, so the mean torque is only 0.36 N m — and it falls to zero twice per revolution, at θ = 0 and θ = 180°. Worse, past θ = 0 the sin θ changes sign and the torque starts braking the rotation. A commutator cures both faults by reversing the current at each dead spot, which flips μ and keeps the torque one way round; real machines then add many windings at staggered angles and shaped pole pieces, so some coil is always near its peak and the delivered torque smooths out.

Every step above leaned on B being the same at every point of the loop. Near a real magnet it is not. The forces on opposite sides then no longer cancel, and an expansion keeps the leading difference: for a small dipole in a field varying along x, Fₓ = μₓ (dBₓ/dx). A dipole with μ = 0.10 A m² sitting where the field climbs at 15 T m⁻¹ feels 1.5 N — a real force, pulling it toward the stronger field when it is aligned and pushing it away when it is reversed. Two consequences follow. A magnet can lift a nail because the nail's induced moment lines up first and is then hauled up the gradient. And the torque returns as the leading effect whenever the loop is small enough that B barely changes across it, which is why τ = μ × B alone describes a compass needle in the Earth's field.