Torque on current loop

Hold a rectangular loop of wire carrying a current inside a uniform magnetic field and something interesting happens: the two sides perpendicular to the field feel equal and opposite forces, but because those forces act along different lines they don't cancel — they form a couple. The loop twists. It keeps twisting until its plane turns perpendicular to the field, at which point the forces still exist but no longer have any turning leverage. This one idea, that a current loop in a magnetic field experiences a torque τ=NBIAsin⁡θ\tau = NBIA\sin\theta that vanishes only at special orientations, is the working principle behind the moving-coil galvanometer and every electric motor you have ever used. This section builds the force-couple picture into the vector law τ⃗=m⃗×B⃗\vec\tau = \vec m\times\vec B and gives you the two orientation cases examiners love to test.

The core insight

A current-carrying loop of magnetic moment m⃗=NIA⃗\vec m = NI\vec A placed in a field B⃗\vec B feels a torque τ⃗=m⃗×B⃗\vec\tau = \vec m\times\vec B, maximum when the loop's plane is parallel to BB and zero when it is perpendicular to BB.

The core idea

Where the couple comes from. Take a rectangular loop of sides aa (length) and bb (breadth), carrying current II, with its normal n^\hat n making angle θ\theta with a uniform field B⃗\vec B. The two sides of length bb, which lie perpendicular to the component of B⃗\vec B in the plane of the loop, each feel a force of magnitude F=BIbF = BIb from F⃗=Il⃗×B⃗\vec F = I\vec l\times\vec B, and these two forces point in opposite directions. They are equal, opposite, and offset — the textbook definition of a couple — so they produce a net torque even though the net force on the loop is zero.

The scalar result. Working through the perpendicular distance between the two force lines (which shrinks as sin⁡θ\sin\theta falls), the torque magnitude on a single-turn loop of area A=abA = ab comes out to:

τ=BIAsin⁡θ\tau = BIA\sin\theta

where BB is the field strength (tesla), II is the current (ampere), AA is the loop's area (m²), and θ\theta is the angle between the field B⃗\vec B and the normal n^\hat n to the loop. For a coil of NN identical turns, each turn contributes the same couple, so the torques simply add:

τ=NBIAsin⁡θ\tau = NBIA\sin\theta

Packaging it as a dipole moment. It is convenient to define the loop's magnetic dipole moment:

m⃗=NIA⃗\vec m = NI\vec A

where A⃗\vec A is a vector of magnitude AA directed along n^\hat n (found from the current's sense of circulation via the right-hand rule). Substituting turns the scalar result into the compact vector law:

τ⃗=m⃗×B⃗\vec\tau = \vec m\times\vec B

which has exactly the same form as the torque on an electric dipole in a uniform electric field, τ⃗dipole=p⃗×E⃗\vec\tau_{\text{dipole}} = \vec p\times\vec E — the same mathematics, a different physical dipole. Two orientations matter most for exams:

  • θ=90°\theta = 90° (m⃗⊥B⃗\vec m \perp \vec B, plane of loop parallel to B⃗\vec B): torque is maximum, τ=NBIA\tau = NBIA.
  • θ=0°\theta = 0° or 180°180° (m⃗∥B⃗\vec m \parallel \vec B, plane of loop perpendicular to B⃗\vec B): torque is zero — θ=0°\theta = 0° is a stable equilibrium (minimum energy), θ=180°\theta = 180° is unstable (maximum energy). Both give τ=0\tau = 0, but only one is stable, and this distinction is a favourite trap.
Figure
A rectangular current loop in a uniform magnetic field with the two force arrows on opposite sides forming a couple and the normal vector at angle theta to the field
Force couple on a current loop tilted at angle theta to B B a b F = BIb F = BIb n̂ (m) θ Equal opposite forces on the b-sides, offset by a sinθ, form the couple giving torque NBIA sinθ
Derivation ATorque on a current loop, from the force on each side

The goal here is to see where τ=NBIAsin⁡θ\tau = NBIA\sin\theta comes from, so that the formula is a consequence of the force on a current-carrying wire rather than something to memorise. We take a single rectangular loop of length ll and breadth bb, carrying current II, sitting in a uniform field BB, and add up the forces on its sides.

Step 1. Consider the two sides of length ll that are perpendicular to the field. Each experiences a force F=BIlF = BIl from the motor rule, and the two forces are equal in magnitude but opposite in direction — they form a couple (the other two sides give forces that act along the same line and cancel with no turning effect).

Step 2. The two forces are separated by the breadth bb, so the couple's torque is force times perpendicular separation. When the plane of the loop is parallel to BB, that separation is the full bb, giving τ=BIl×b=BI(lb)=BIA\tau = BIl \times b = BI(lb) = BIA, where A=lbA = lb is the loop's area.

Step 3. As the loop turns so its normal makes angle θ\theta with BB, the perpendicular distance between the forces shrinks to bsin⁡θb\sin\theta, so τ=BIAsin⁡θ\tau = BIA\sin\theta. For a coil of NN turns every turn contributes equally, giving the general result τ=NBIAsin⁡θ\tau = NBIA\sin\theta, maximum when the plane is parallel to BB (θ=90∘\theta = 90^\circ) and zero when the plane is perpendicular to BB (θ=0\theta = 0).

Worked example

A rectangular coil of 5050 turns, with dimensions 0.1 m×0.05 m0.1\ \text{m}\times0.05\ \text{m}, carries a current of 2 A2\ \text{A} and sits in a uniform field of 0.3 T0.3\ \text{T} with the plane of the coil parallel to the field. Find the torque, then find the torque if the coil is instead rotated so its plane makes 30°30° with the field.

Step 1. Read the geometry: "plane of coil parallel to B⃗\vec B" means the normal n^\hat n (and hence m⃗\vec m) is perpendicular to B⃗\vec B, so θ=90°\theta = 90° and sin⁡θ=1\sin\theta = 1 — this is the maximum-torque case.

Step 2. Find the area: A=0.1×0.05=5×10−3 m2A = 0.1\times0.05 = 5\times10^{-3}\ \text{m}^2.

Step 3. Apply τ=NBIAsin⁡θ\tau = NBIA\sin\theta with N=50N = 50, B=0.3 TB = 0.3\ \text{T}, I=2 AI = 2\ \text{A}, sin⁡θ=1\sin\theta = 1: τ=50×0.3×2×5×10−3=0.15 N ⋅ m\tau = 50\times0.3\times2\times5\times10^{-3} = 0.15\ \text{N}\,\cdot\,\text{m}.

Step 4. Now handle the rotated case. If the coil's plane makes 30°30° with B⃗\vec B, the normal makes θ=90°−30°=60°\theta = 90° - 30° = 60° with B⃗\vec B — the angle in the torque formula is always the one between m⃗\vec m (the normal) and B⃗\vec B, not the one between the plane and B⃗\vec B.

Step 5. Compute sin⁡60°=0.866\sin60° = 0.866, so τ=50×0.3×2×5×10−3×0.866=0.15×0.866=0.130 N ⋅ m\tau = 50\times0.3\times2\times5\times10^{-3}\times0.866 = 0.15\times0.866 = 0.130\ \text{N}\,\cdot\,\text{m}.

The torque on the coil is 0.15 N ⋅ m\boxed{0.15\ \text{N}\,\cdot\,\text{m}} when its plane is parallel to B⃗\vec B, falling to 0.130 N ⋅ m\boxed{0.130\ \text{N}\,\cdot\,\text{m}} when the plane is tilted to 30°30° from B⃗\vec B — smaller because the loop is now closer to the zero-torque equilibrium orientation.

Common traps

  • Confusing the angle θ\theta in τ=NBIAsin⁡θ\tau = NBIA\sin\theta (between m⃗\vec m and B⃗\vec B) with the angle between the plane of the loop and B⃗\vec B — these are complementary (θplane=90°−θm⃗,B⃗\theta_{\text{plane}} = 90° - \theta_{\vec m,\vec B}), so mixing them up silently swaps sin⁡θ\sin\theta for cos⁡θ\cos\theta.
  • Forgetting the factor NN for a multi-turn coil — a common source of a factor-of-NN numerical error.
  • Treating both zero-torque orientations as equivalent: m⃗∥B⃗\vec m \parallel \vec B is stable equilibrium, m⃗\vec m antiparallel is unstable — both give τ=0\tau = 0, but the loop only settles at the stable one.
  • Assigning the direction of m⃗\vec m arbitrarily instead of getting it from the right-hand rule applied to the current's actual sense of circulation.

Ask the tutor: Not sure whether to use sin⁡θ\sin\theta or cos⁡θ\cos\theta for a given loop orientation, or want the force-couple geometry walked through with a labelled diagram? Ask the AI tutor.

Why does the loop stop rotating exactly when its plane is perpendicular to B, and not just slow down gradually?

The torque itself falls smoothly as sin⁡θ\sin\theta shrinks, so the loop does slow its angular acceleration gradually — but the point where it stops experiencing any turning force at all is exactly θ=0\theta = 0, where sin⁡θ=0\sin\theta = 0 makes τ=0\tau = 0 identically. In a real coil with some damping (as in a galvanometer), the loop settles into this zero-torque, stable-equilibrium orientation rather than oscillating forever. Without damping it would actually overshoot and oscillate about that equilibrium, much like a pendulum about its lowest point.

Is the angle theta in the torque formula measured from the plane of the loop or from the normal to the loop?

Always from the normal n^\hat n, which is the same direction as m⃗\vec m. This is the single most common source of error: many students instinctively measure the angle between the plane and B⃗\vec B, but τ=NBIAsin⁡θ\tau = NBIA\sin\theta needs θ\theta between m⃗\vec m and B⃗\vec B. If a question gives you the angle to the plane, subtract it from 90°90° before substituting, or equivalently swap sin⁡\sin for cos⁡\cos of the given angle.

How is this torque formula related to the moving-coil galvanometer?

The galvanometer places a current-carrying coil inside a magnetic field and lets this exact torque τ=NBIAsin⁡θ\tau = NBIA\sin\theta rotate the coil against a restoring spring; the coil settles where the magnetic torque balances the spring's torque, and the deflection angle becomes a readout of the current II. Practical galvanometers use a radial field so θ\theta stays close to 90°90° throughout the swing, keeping sin⁡θ≈1\sin\theta \approx 1 and making the deflection linear in II rather than following the full sin⁡θ\sin\theta curve derived here.

Before this

Where this leads