Force on a current-carrying conductor

A current-carrying conductor creates its own magnetic field, and an external magnetic field acts on every moving charge carrier inside it. The combined effect is the Ampère force applied to the conductor.

For a straight segment of length $l$ in a uniform field of induction $B$ the magnitude of the force is

$$F = B I l \sin\alpha,$$

where $\alpha$ is the angle between the current direction and the vector $\vec B$. In vector form $\vec F = I\,\vec l \times \vec B$, so the force is always perpendicular both to the conductor and to the magnetic field.

Its direction is conveniently found with the left-hand rule: fingers along the current, field lines entering the palm, and the extended thumb gives the force. If the conductor is parallel to the field ($\alpha = 0$), then $\sin\alpha = 0$ and the force is zero; it is largest at $\alpha = 90^\circ$, where $F = BIl$.

A straight conductor of length $l = 0.20$ m carries a current $I = 5.0$ A and is placed perpendicular to a uniform magnetic field of induction $B = 0.40$ T. Find the magnitude of the Ampère force and state its direction if the current points east and the field points vertically upward.

A conductor of length $l = 0.15$ m carries a current $I = 4.0$ A and lies in a uniform field $B = 0.25$ T at an angle of $30^\circ$ to the field lines.

(a) Find the Ampère force.
(b) By what factor does the force change if the conductor is turned to $\alpha = 90^\circ$ while the current and the field stay the same?

A horizontal straight conductor of length $l = 0.20$ m and mass $m = 12$ g is held in a uniform horizontal magnetic field; the current $I = 3.0$ A is perpendicular to the field. What must the induction $B$ be for the Ampère force to balance the weight? State the mutual arrangement of $\vec B$, the current and the force. Take $g = 9.8$ m/s$^2$.

Self-study plan in a simulation (no experiments with 220 V mains and no strong permanent magnets outside a simulation).

1. Open a simulation of a current-carrying conductor in a magnetic field (for example PhET or an equivalent virtual Lorentz/Ampère force bench).
2. Set $B = 0.20$ T, conductor length $l = 0.20$ m and angle $\alpha = 90^\circ$. Measure the force for currents $I = 1$ A, $2$ A, $3$ A. Plot $F(I)$ and check that it is a straight line through the origin: $F \propto I$.
3. At a fixed current $I = 2$ A, halve $B$ and check that the force also halves ($F \propto B$).
4. Rotate the conductor, setting $\alpha = 0^\circ, 30^\circ, 60^\circ, 90^\circ$, and fill a table of $F$ versus $\sin\alpha$. Check that the points lie on a straight line and that the force becomes zero at $\alpha = 0^\circ$.
5. For each configuration, first predict the force direction with the left-hand rule, then verify it with the force arrow in the simulation; separately reverse the current and reverse the field.
6. Wrap up: write in your notebook the three proportionalities — $F \propto I$, $F \propto B$, $F \propto \sin\alpha$ — and the resulting formula $F = BIl\sin\alpha$.

Show that the Ampère force is perpendicular both to the current and to the magnetic field, and how its magnitude depends on the angle $\alpha$ between the current direction and the vector $\vec B$.

Diagram: a straight conductor at an angle $\alpha$ to parallel field lines, with arrows for $\vec B$ along the lines, $\vec I$ along the conductor, and $\vec F$ perpendicular to both. Beside it, a graph of $F$ versus $\sin\alpha$: a straight line through the origin, reaching its largest value $F = BIl$ at $\sin\alpha = 1$.

When a rider opens the throttle, a section of copper conductor in the hub motor of length $L=0.12\,\text{m}$ carries current $I=15\,\text{A}$ through a magnetic field $B=0.30\,\text{T}$ at $\theta=90^\circ$ to the conductor. The force on that section is $F=BIL\sin\theta=0.30\cdot15\cdot0.12\cdot1=0.54\,\text{N}$. Many such conductor sections around the rotor add their forces and make the wheel turn.

This uses the Ampere force law $F=BIL\sin\theta$ with the left-hand rule: the $90^\circ$ angle gives the maximum force, and the force direction is perpendicular to both $\vec B$ and the current, so the sum of forces produces torque on the rotor.

In a phone loudspeaker the voice coil sits in a radial magnetic field $B=0.45\,\text{T}$. An audio current $I=0.08\,\text{A}$ flows through an effective conductor length $L=1.6\,\text{m}$ in the gap, nearly perpendicular to the field ($\theta=90^\circ$). The force on the coil is $F=BIL\sin\theta=0.45\cdot0.08\cdot1.6=0.0576\,\text{N}$, about $0.058\,\text{N}$, and it pushes the coil and cone.

This is the Ampere force on the coil: an alternating audio current reverses the current direction, so the force direction reverses each half-cycle and vibrates the cone at the audio frequency; $F=BIL\sin\theta$ gives the instantaneous force magnitude.

A tram traction motor has a stator field $B=0.60\,\text{T}$. One active conductor section in the rotor of length $L=0.20\,\text{m}$ carries current $I=120\,\text{A}$ at $\theta=90^\circ$ to the field. The force on it is $F=BIL\sin\theta=0.60\cdot120\cdot0.20=14.4\,\text{N}$; hundreds of such sections create the torque that starts the tram.

This applies $F=BIL\sin\theta$ and the left-hand rule: the stator electromagnet supplies $\vec B$, the rotor current $I$ is perpendicular to it in the active sides, and the resulting force on each side forms a couple that rotates the armature.