Force on Current-Carrying Conductors

Take a segment of length L and cross-section A holding n mobile carriers per cubic metre, each of charge q drifting at speed v_d. The segment contains N = n A L carriers and each feels q v_d × B, so the total is F = n A L (q v_d × B). Regroup the factors: n q v_d A is exactly the current I, and the length measured along the conventional current is the vector L, so F = I L × B. Nothing about the carriers survives. Copper has n = 8.5 × 10²⁸ m⁻³, so a 1.0 mm² wire carrying 3.0 A drifts at v_d = I/(nqA) = 0.22 mm s⁻¹, and in a 0.40 T field one electron feels about 1.4 × 10⁻²³ N — but 0.25 m of that wire holds 2.1 × 10²² of them, and the product is 0.30 N. The sign looks after itself: make the carriers negative and v_d reverses too, so the product q v_d is unchanged. That is why the force cannot tell you whether a metal conducts by electrons or by positive holes.

F = I L × B is a cross product, so the force is perpendicular to the wire and to the field at once, and its direction comes from the right-hand rule: fingers along the conventional current, curl them into B, thumb along F. A wire carrying current due east through a field pointing due north is pushed vertically upward. The magnitude carries sin θ, and only the component of B perpendicular to the wire contributes, so F = I L B_⊥. Swing the 0.25 m wire of the definition to 30° from the field and its 0.30 N drops to 3.0 × 0.25 × 0.40 × sin 30° = 0.15 N; lay it along B and dL × B vanishes element by element, leaving no force at all. Note also that F never has a component along the wire. A magnetic field can push a conductor sideways or bend it, but it cannot drive current along it.

If B is the same at every element it comes outside the integral: F = I (∫ dL) × B. The integral that is left adds up tiny displacement vectors, so it is simply the net displacement from the start of the wire to its end. Check that on a semicircle of radius R, parameterised as (R cos θ, R sin θ) with θ running from 0 to π, so that dL = (−R sin θ, R cos θ) dθ. The y-component integrates to R[sin θ] between 0 and π, which is zero by symmetry; the x-component gives R[cos θ] between 0 and π, which is −2R. So ∫dL = −2R x̂, the chord, exactly as claimed. With R = 0.12 m, I = 4.0 A and B = 0.30 T out of the page, the wire is πR = 0.377 m long but the force is I(2R)B = 4.0 × 0.24 × 0.30 = 0.288 N, not the 0.45 N the arc length would suggest.

Close the path and the start and the finish coincide, so ∮ dL = 0 and the net force is exactly zero however the loop is shaped or oriented. This is not a claim that nothing is pushed. On a rectangular loop the opposite sides carry current in opposite directions, so they are pushed in opposite directions: the pair either stretches or squeezes the loop, and when the loop is tilted the two forces act along different lines and deliver a couple. That torque is the next topic; what belongs here is that the translation vanishes, which is why a loop between flat pole faces spins on its axis instead of flying out from between them. The zero depends entirely on B being identical at every element. Hold the loop near one pole, where the field spreads and weakens, and the cancellation is incomplete — that leftover is the force that drags a magnetic dipole toward a stronger field.

Once B differs from element to element it cannot be factored out, and the endpoint shortcut dies with it. Set the integral up in whatever coordinate B depends on. Let a wire run along x from 0 to 0.50 m carrying I = 6.0 A in the +x direction through a field B(x) = B₀(1 + x/d) ẑ, with B₀ = 0.20 T and d = 0.50 m, so the field doubles across the wire. Each element gives dF = I dx (x̂ × ẑ) B(x), and since x̂ × ẑ = −ŷ every element pushes the same way, so only magnitudes add: F = I ∫ B₀(1 + x/d) dx from 0 to 0.50 m = I B₀ [x + x²/(2d)] from 0 to 0.50 m = 6.0 × 0.20 × 0.75 = 0.90 N. That matches using the mean field 0.30 T, which is what the integral computes. Designers dodge the integral instead: a loudspeaker puts its coil in a radial gap where B has one magnitude everywhere and always sits perpendicular to the wire, so F = N I L B holds with no integration and no dependence on how far the coil has travelled.

The magnetic force acts on the carriers, not the metal. The carriers drift sideways for an instant, pile up against one face of the conductor, and the transverse electric field that builds up pushes back on them while pulling on the fixed ions. Within nanoseconds the sideways drift stops and the whole force I L × B is carried by the lattice — which is why a wire hung between two magnets swings. That same transverse field is what the Hall effect measures. The energy bookkeeping is stranger. A magnetic force on a charge is always perpendicular to that charge's velocity, so it does no work on the carriers — yet a motor plainly delivers work. Once the wire moves at speed u, each carrier's velocity is v_d along the wire plus u sideways; the magnetic force then has one component doing positive work on the wire and another opposing the drift, and the two cancel exactly. That second component is not itself an emf — it is the force per unit charge u B along the wire, and the back emf is what it integrates to over the length, ε = B ℓ u. The mechanical energy comes from the supply driving current against that back emf; the field only brokers it. A 0.10 m rail carrying 15 A across 0.25 T is pushed with I L B = 0.375 N.