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How does a current create a magnetic field, and what force does a field exert on a current?

Topic 12.3 Magnetism and Current-Carrying Wires: relate currents to the magnetic fields they create and the forces they experience in a field.

A focused answer to AP Physics 2 Topic 12.3, covering the magnetic field around a straight current-carrying wire, the right-hand rule for its direction, the field of a solenoid, the force on a current-carrying wire F = BIL sin theta, and the forces between parallel currents, with full worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

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Quick answer

A current creates a magnetic field. Around a long straight wire the field circles the wire, $B = \dfrac{\mu_0 I}{2\pi r}$ , falling off as $1/r$ ; its direction is given by the right-hand rule (thumb along the current, curled fingers show the field). A solenoid (coil) produces a nearly uniform field inside, like a bar magnet, which is the basis of an electromagnet. A magnetic field also exerts a force on a current-carrying wire: $F = BIL\sin\theta$ , where $I$ is the current, $L$ the length in the field and $\theta$ the angle between the current and the field (maximum when perpendicular, zero when parallel). The direction follows the right-hand rule. Two parallel wires interact through each other's fields: same-direction currents attract, opposite-direction currents repel. These effects, current makes a field, and a field pushes a current, are the working principle of motors, electromagnets and loudspeakers.

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What this topic is asking
A current creates a magnetic field
The force on a current-carrying wire
Forces between parallel wires
Try this

What this topic is asking

The College Board (Topic 12.3) wants you to relate currents to the magnetic fields they create (a straight wire and a solenoid), and to calculate the force on a current-carrying wire in a field, $F = BIL\sin\theta$ , including the forces between parallel currents.

A current creates a magnetic field

This is the discovery that unified electricity and magnetism: a current is a source of magnetic field. Around a single straight wire the field forms circles, weakening as $1/r$ with distance (gentler than a point charge's $1/r^2$ ). Winding the wire into a coil adds the loops' fields together inside, making a strong, uniform field, an electromagnet you can switch on and off. The right-hand rule (thumb along current, fingers curl with the field) gives the direction in every case.

The force on a current-carrying wire

This force is just the magnetic force on moving charges (Topic 12.2) added up over all the charges flowing in the wire: $F = BIL\sin\theta$ is the wire-current version of $F = qvB\sin\theta$ . The same angle dependence applies, full force when the current cuts across the field, none when it runs along it. This force is what turns a motor: a current loop in a field feels forces that make it rotate.

Forces between parallel wires

Because each wire creates a field and each wire in a field feels a force, two parallel wires exert forces on each other. Applying the field-of-a-wire and the force-on-a-wire together shows that currents in the same direction attract, and currents in opposite directions repel, the opposite of the rule for like and unlike electric charges, so worth memorizing carefully. The strategic role of this topic is that it closes the loop between current and magnetism: a current is a source of magnetic field (extending Topic 12.1), and a magnetic field exerts a force on a current (extending Topic 12.2). Together these explain electromagnets, motors, loudspeakers and the magnetic interaction of circuits, and they set up the final idea of the unit, that a changing magnetic field induces a current (Topic 12.4).

Force on a wire in a magnetic field

A $0.50$ m wire carries $6.0$ A at $30$ degrees to a $0.40$ T magnetic field. Find the force on the wire, then the force if it is rotated perpendicular to the field.

step 1 Force at 30 degrees

$F = BIL\sin\theta = (0.40)(6.0)(0.50)\sin 30^\circ = (0.40)(6.0)(0.50)(0.50) = 0.60$ N.

step 2 Force when perpendicular

At $\theta = 90^\circ$ , $\sin 90^\circ = 1$ : $F = (0.40)(6.0)(0.50)(1) = 1.2$ N.

step 3 Compare

Rotating from $30^\circ$ to $90^\circ$ doubles the force (since $\sin 90^\circ / \sin 30^\circ = 1/0.5 = 2$ ), confirming the force is greatest perpendicular to the field.

step 4 Interpret

The wire feels the most force when it cuts straight across the field lines, and none when aligned with them, the same angle rule as for a single moving charge.

Try this

Q1. State whether two parallel wires carrying current in the same direction attract or repel. [1 point]

Cue. They attract.

Q2. A $0.20$ m wire carries $5.0$ A perpendicular to a $0.30$ T field. Calculate the force on it. [2 points]

Cue. $F = BIL\sin 90^\circ = (0.30)(5.0)(0.20)(1) = 0.30$ N.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). A straight wire of length

0.30

m carries a current of

4.0

A perpendicular to a uniform magnetic field of

0.25

T. (a) Calculate the magnetic force on the wire. (b) Describe how to find the field direction around a long straight current-carrying wire. (c) Two long parallel wires carry currents in the same direction. State and justify whether they attract or repel.

Show worked answer →

A 7-point FRQ on current-carrying wires.

(a) Force (2 points): $F = BIL\sin\theta = (0.25)(4.0)(0.30)\sin 90^\circ = 0.30$ N.
(b) Field direction (2 points): use the right-hand rule for a wire: point the thumb along the conventional current, and the curled fingers give the circular direction of the magnetic field around the wire.
(c) Parallel currents (3 points): they attract. Each wire sits in the magnetic field of the other; applying the force law shows that currents in the same direction experience attractive forces (and opposite currents repel).

Markers reward the force formula, the right-hand rule for the field around a wire, and the attraction of same-direction currents.

AP 2023 (style)1 marksSection I (multiple choice). The magnetic field at a distance

r

from a long straight current-carrying wire depends on

r

how? (A) it is independent of

r

(B) it is proportional to

r

1/r

(D) it is proportional to

1/r^2

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the field of a straight wire. The answer is (C).

The magnetic field around a long straight wire is $B = \dfrac{\mu_0 I}{2\pi r}$ , which falls off as $1/r$ (not $1/r^2$ like a point-charge field). Doubling the distance halves the field. The trap is (D): the wire's field is inverse in $r$ , not inverse-square.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)