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What force does a magnetic field exert on a moving charge, and why does it make charges move in circles?

Topic 12.2 Magnetism and Moving Charges: calculate the magnetic force on a moving charge and describe the resulting circular motion.

A focused answer to AP Physics 2 Topic 12.2, covering the magnetic force on a moving charge F = qvB sin theta, the right-hand rule for direction, why the force does no work, the resulting circular motion and its radius, and the dependence on the angle to the field, with full worked examples.

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

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Quick answer

A magnetic field exerts a force on a moving charge: $F = qvB\sin\theta$ , where $q$ is the charge, $v$ its speed, $B$ the field, and $\theta$ the angle between the velocity and the field. The force is zero when the charge moves parallel to the field ( $\theta = 0$ ) and maximum when it moves perpendicular ( $\theta = 90^\circ$ ). The direction is perpendicular to both the velocity and the field, given by the right-hand rule (reverse it for a negative charge). Because the force is always perpendicular to the velocity, it does no work: it changes the charge's direction but not its speed or kinetic energy. A charge moving perpendicular to a uniform field therefore travels in a circle, with the magnetic force providing the centripetal force, $qvB = \dfrac{mv^2}{r}$ , giving radius $r = \dfrac{mv}{qB}$ . A stationary charge feels no magnetic force; only moving charges do.

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What this topic is asking
The magnetic force on a moving charge
Direction and why no work is done
Circular motion in a magnetic field
Try this

What this topic is asking

The College Board (Topic 12.2) wants you to calculate the magnetic force on a moving charge, $F = qvB\sin\theta$ , find its direction with the right-hand rule, explain why it does no work, and describe the resulting circular motion.

The magnetic force on a moving charge

Unlike the electric force, the magnetic force acts only on moving charges and depends on the angle between the motion and the field. The $\sin\theta$ factor means a charge gliding along the field lines feels nothing, while one cutting straight across them feels the full force $qvB$ . This angle dependence is tested directly: parallel motion gives zero force, perpendicular gives the maximum.

Direction and why no work is done

The perpendicular nature has a profound consequence. A force at right angles to the velocity does zero work ( $W = Fd\cos 90^\circ = 0$ ), so the magnetic force can never speed up or slow down a charge, only bend its path. This is why magnetic fields steer charged particles, in mass spectrometers, particle accelerators and old television tubes, without adding energy. The right-hand rule gives the direction; remember to flip it for electrons and other negative charges.

Circular motion in a magnetic field

When a charge moves perpendicular to a uniform field, the magnetic force has constant magnitude and always points toward a center, exactly the condition for uniform circular motion. Setting the magnetic force equal to the centripetal force gives the radius:

qvB = \dfrac{mv^2}{r} \quad\Rightarrow\quad r = \dfrac{mv}{qB}

The radius grows with the particle's momentum ( $mv$ ) and shrinks with stronger fields or larger charge: faster or heavier particles circle wider, stronger fields curl them tighter. The strategic value of this topic is that it connects magnetism to the circular-motion and energy ideas of mechanics: the magnetic force supplies the centripetal force (like tension or gravity did before), yet does no work, so the speed and kinetic energy stay constant. This same force, summed over the moving charges in a wire, gives the force on a current-carrying conductor in Topic 12.3, and the relative motion of charges and fields underlies the induction of Topic 12.4.

An electron circling in a magnetic field

An electron (charge magnitude $1.6 \times 10^{-19}$ C, mass $9.11 \times 10^{-31}$ kg) moves at $3.0 \times 10^6$ m/s perpendicular to a $0.20$ T field. Find the force and the radius of its path.

step 1 Magnetic force

$F = qvB\sin 90^\circ = (1.6 \times 10^{-19})(3.0 \times 10^6)(0.20)(1) = 9.6 \times 10^{-14}$ N.

step 2 Set magnetic force equal to centripetal force

$qvB = \dfrac{mv^2}{r}$ , so $r = \dfrac{mv}{qB}$ .

step 3 Solve for the radius

$r = \dfrac{(9.11 \times 10^{-31})(3.0 \times 10^6)}{(1.6 \times 10^{-19})(0.20)} = \dfrac{2.73 \times 10^{-24}}{3.2 \times 10^{-20}} = 8.5 \times 10^{-5}$ m.

step 4 Interpret

The light electron circles in a tiny path ( $85$ micrometers). The force bends its motion into a circle but, being perpendicular to the velocity, never changes its speed.

Try this

Q1. State the magnetic force on a charge moving parallel to a magnetic field. [1 point]

Cue. Zero ( $\sin 0 = 0$ ).

Q2. Explain why a magnetic force cannot change a charged particle's speed. [1 point]

Cue. It is always perpendicular to the velocity, so it does no work; it changes direction but not speed.

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 proton (charge

1.6 \times 10^{-19}

C, mass

1.67 \times 10^{-27}

kg) moves at

2.0 \times 10^6

m/s perpendicular to a uniform magnetic field of

0.50

T. (a) Calculate the magnitude of the magnetic force on the proton. (b) Explain why this force does no work on the proton. (c) Calculate the radius of the proton's circular path.

Show worked answer →

A 7-point FRQ on the magnetic force on a moving charge.

(a) Force (2 points): $F = qvB\sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)\sin 90^\circ = 1.6 \times 10^{-13}$ N.
(b) No work (2 points): the magnetic force is always perpendicular to the velocity, so it does no work ( $W = Fd\cos 90^\circ = 0$ ); it changes the direction of motion but not the speed or kinetic energy.
(c) Radius (3 points): the magnetic force provides the centripetal force, $qvB = \dfrac{mv^2}{r}$ , so $r = \dfrac{mv}{qB} = \dfrac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)} = \dfrac{3.34 \times 10^{-21}}{8.0 \times 10^{-20}} = 0.042$ m.

Markers reward the force formula, the perpendicular/no-work argument, and equating the magnetic force to the centripetal force for the radius.

AP 2023 (style)1 marksSection I (multiple choice). A charged particle moves parallel to a uniform magnetic field. What is the magnetic force on it? (A) maximum (B) zero (C) perpendicular to the field (D) equal to qvB. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the angle dependence. The answer is (B).

The magnetic force is $F = qvB\sin\theta$ . When the velocity is parallel to the field, $\theta = 0$ and $\sin 0 = 0$ , so the force is zero. The force is greatest when the motion is perpendicular to the field ( $\theta = 90^\circ$ ). The trap is (D): $qvB$ is the maximum, reached only at $90^\circ$ , not for parallel motion.

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Sources & how we know this

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