Forces on Current-Carrying Conductors
Forces on Current-Carrying Conductors
A current-carrying wire in a magnetic field experiences a force. This is the principle behind every electric motor, loudspeaker, and electromagnetic relay in the world. The force arises because current is simply moving charges, and each charge in the wire experiences the Lorentz force. When we sum the forces over all the charge carriers in a length of wire, we get a simple and powerful formula: F equals I L B sine theta, where I is the current, L is the length of wire in the field, B is the magnetic field strength, and theta is the angle between the wire and the field.
This formula is the foundation of electromechanical energy conversion. In a motor, current-carrying coils sit in a magnetic field. The force on each segment of wire creates a torque that spins the rotor. The torque on a rectangular current loop in a uniform field is tau equals N I A B sine alpha, where N is the number of turns, A is the area of the loop, and alpha is the angle between the loop's normal and the field. This is why motors have many turns of wire and strong magnets — both increase the torque.
Two parallel wires carrying current also exert forces on each other. Each wire creates a magnetic field, and the other wire sits in that field. Parallel currents attract; antiparallel currents repel. This force between wires was historically used to define the ampere: one ampere is the current that produces a force of 2 times 10 to the negative 7 newtons per meter between two parallel wires one meter apart.
The mathematics in this lesson combines algebra, trigonometry, and the geometry of forces. You will calculate forces on wires, torques on loops, and forces between parallel conductors — all with straightforward formulas that connect directly to real-world devices.
A straight wire of length $L$ carrying current $I$ in a magnetic field $\vec{B}$:
$$F = BIL\sin\theta$$
where $\theta$ is the angle between the wire and the field. Direction: right-hand rule (fingers along current, curl toward $\vec{B}$, thumb is $\vec{F}$).
A rectangular loop with $N$ turns, area $A$, current $I$ in field $B$:
$$\tau = NIAB\sin\alpha$$
where $\alpha$ is the angle between the loop's normal and $\vec{B}$. Maximum torque occurs when $\alpha = 90°$.
The force per unit length between two parallel wires separated by distance $d$:
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
Parallel currents (same direction) → attract. Antiparallel → repel.
A 50-cm wire carrying 8 A lies perpendicular to a 0.4 T magnetic field. Find the force on the wire.
Step 1: Since the wire is perpendicular, $\theta = 90°$ and $\sin\theta = 1$.
$$F = BIL = (0.4)(8)(0.50) = 1.6 \text{ N}$$
This is the force that makes electric motors turn.
A motor coil has 200 turns, area 0.02 m², carrying 3 A in a 0.5 T field. Find the maximum torque.
Step 1: Maximum torque at $\alpha = 90°$:
$$\tau_{\max} = NIAB = (200)(3)(0.02)(0.5) = 6.0 \text{ N·m}$$
Two parallel wires 5 cm apart carry currents of 10 A and 15 A. Find the force per meter between them.
Step 1: Apply the parallel wire force formula.
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(10)(15)}{2\pi (0.05)}$$
Step 2: Simplify.
$$\frac{F}{L} = \frac{4 \times 10^{-7} \times 150}{0.10} = \frac{6 \times 10^{-5}}{0.10} = 6 \times 10^{-4} \text{ N/m} = 0.6 \text{ mN/m}$$
Practice Problems
Show Answer Key
1. $F = (0.6)(12)(0.30)\sin60° = 2.16 \times 0.866 = 1.87$ N
2. $\tau = (500)(2)(0.05)(0.3) = 15.0$ N·m
3. $F/L = \mu_0(20)(20)/(2\pi \times 0.08) = 1.0 \times 10^{-3}$ N/m $= 1.0$ mN/m
4. $\sin\theta = 0.5 \implies \theta = 30°$
5. $B = F/(IL) = 1/(5 \times 1) = 0.2$ T
6. $A = \pi(0.01)^2 = 3.14 \times 10^{-4}$ m². $\tau = (50)(0.5)(3.14\times10^{-4})(0.8) = 6.28 \times 10^{-3}$ N·m