Magnetic Force on a Current-Carrying Wire

F=BILsin⁡θF = B I L \sin\theta

Worked example: 10 A in 2 m of wire across 0.5 T → 10 N — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Magnets at work →

Grade 11Grade 11 Physics

Wires in the field →

Grade 12Grade 12 Physics

Wires and forces →

UniversityCircuits & Electrical Power

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning.

See your Report Card
Compete with your friends
share your results
Learning zone

Magnetic Force on a Current-Carrying Wire explained

IFBL

This is the force on one moving charge, F=qvBsin⁡θF = qvB\sin\theta, added up over all the charges in a length of wire. A current II means charge crossing at II coulombs per second, so a length LL of conductor holds a quantity of moving charge whose product with its drift speed is exactly ILIL — the individually feeble pushes on perhaps 102210^{22} slowly drifting electrons, collected by the metal lattice and delivered to the wire as a whole. That is why F=BILsin⁡θF = BIL\sin\theta contains no reference to how many carriers there are or how fast they move: those two factors always multiply out to the current. The sine handles orientation, peaking when the wire lies across the field and vanishing when it lies along it, since a charge coasting parallel to a field feels nothing.

A 0.25 m length of wire carrying 8 A across a 0.4 T field at right angles feels F=0.4×8×0.25=0.8 NF = 0.4 \times 8 \times 0.25 = 0.8\ \text{N} — about the weight of a coffee mug, from a single conductor. Multiply by a few hundred turns in an armature and you have the torque of a real motor. Tilt the same wire to 30° from the field and the force drops to 0.8sin⁡30°=0.4 N0.8 \sin 30° = 0.4\ \text{N}, half. The direction is perpendicular to both the wire and the field, given by the right-hand rule, and this sideways push is what motor designers arrange to be a torque.

Faraday demonstrated the effect in 1821 with a wire free to rotate around a magnet dipped in mercury — the first electric motor, built to settle an argument about whether electromagnetism could produce continuous motion. Every motor and every loudspeaker since is the same experiment industrialised, the cone driven by a coil of wire hanging in a permanent magnet's gap with the audio signal as II. Note that this relation and the motional-EMF page are two faces of one thing: push current through a wire in a field and it moves, move a wire in a field and current appears, and a motor and a generator are the same machine run in opposite directions.

Three cautions. The angle θ\theta is measured between the wire and the field, and it cannot be solved for on this page — arcsine cannot tell θ\theta from its supplement 180°−θ180° - \theta, so the calculator returns FF, BB, II or LL but never the angle. LL is the length of conductor actually inside the field, not the length of the wire; a metre of lead-in outside the magnet gap contributes nothing. And a note on the right-hand rule: it works with conventional current, drawn flowing from positive to negative, while the electrons in the copper are travelling the other way. Benjamin Franklin fixed that sign a century before the electron was found, and he fixed it backwards. Both descriptions give the same force in the same direction — negative charge moving left is the same current as positive charge moving right — but if you switch to reasoning about electrons you must switch hands too, and mixing the two is the surest way to get a motor turning the wrong way on paper.

Magnetic Force on a Current-Carrying Wire formula

F=BILsin⁡θF = B I L \sin\theta
Where
  • FF= Magnetic force (N)
  • BB= Magnetic flux density (T)
  • II= Current (A)
  • LL= Wire length in field (m)
  • θ\theta= Angle between wire and B (°)