In AP Physics 2, circular motion describes a charged particle moving at constant speed in a circle when its velocity is perpendicular to a uniform magnetic field; the magnetic force F = qvB acts as the centripetal force, giving radius r = mv/(qB).
Circular motion is what happens when a charged particle enters a uniform magnetic field with its velocity perpendicular to the field. The magnetic force on the particle is F = qvB sin θ, and when θ = 90°, that force is always perpendicular to the velocity. A force that's perpendicular to motion can't speed the particle up or slow it down. It can only turn it. So the particle keeps the same speed but constantly changes direction, tracing a circle.
The magic move on the exam is setting the magnetic force equal to the centripetal force. Write qvB = mv²/r and solve for whatever you need. Most often that's the radius, r = mv/(qB). Notice what this tells you. Faster or heavier particles make bigger circles, while stronger fields or bigger charges make tighter ones. Also, because the force is always perpendicular to displacement, the magnetic force does zero work on the particle. The kinetic energy never changes.
Circular motion lives in Topic 12.2 (Magnetism and Moving Charges) in Unit 12, supporting learning objective 12.2.B, describing the force a magnetic field exerts on moving charged objects. Essential knowledge 12.2.B.2 gives you the force law F_B = qvB sin θ and the right-hand rule for its direction, and circular motion is the payoff. It's where that force law actually produces a predictable path you can calculate. This is also one of the cleanest places the exam blends old and new physics. You're reusing Newton's second law and centripetal acceleration from mechanics, just with a magnetic force supplying the center-pointing pull. Mass spectrometers, charge-to-mass ratio measurements, and particle trajectory problems all run on this one idea.
Keep studying AP® Physics 2 Unit 12
Centripetal acceleration (Unit 12, from mechanics)
Circular motion in a magnetic field is just uniform circular motion from AP Physics 1 wearing a new costume. The centripetal acceleration is still v²/r, but now the magnetic force qvB is what supplies it. Setting qvB = mv²/r is the single most important equation move in this topic.
F_B = qvB sin θ (Unit 12)
This force law is the engine behind the circle. When velocity is exactly perpendicular to the field, sin θ = 1 and the force is maximized and always perpendicular to motion, which is the precise condition that makes the path a perfect circle instead of a curve that speeds up or slows down.
Charge-to-mass ratio (Unit 12)
Rearrange r = mv/(qB) and you get q/m = v/(rB). Measure the radius of a particle's circular path in a known field and you've measured its charge-to-mass ratio. This is exactly how mass spectrometers identify particles, and it's a favorite setup for multi-step problems.
Velocity selector (Unit 12)
A velocity selector uses crossed electric and magnetic fields to let only particles of one speed pass straight through. It's often paired with circular motion in problems. First the selector picks the speed, then the particle enters a pure magnetic field region and curves into a circle whose radius reveals its mass or charge.
Circular motion shows up mostly as quantitative multiple-choice and short FRQ parts built on qvB = mv²/r. Typical stems ask you to solve for one variable given the others, like finding speed from r, m, q, and B. Proportional reasoning is huge here. A classic question doubles the field strength at constant speed and asks what happens to the radius (it halves, since r = mv/(qB)). You'll also see direction questions where you must use the right-hand rule to state that the magnetic force is perpendicular to the velocity, and conceptual checks like computing the force qvB on a particle moving in a circle of given radius. The conceptual trap they love is asking about energy. The magnetic force is always perpendicular to velocity, so it does no work and the speed stays constant. Be ready to say that in a sentence, not just an equation.
A particle only moves in a flat circle when its velocity is entirely perpendicular to the magnetic field. If the velocity has a component along the field, that component feels no force (sin θ = 0 along the field direction), so the particle drifts along the field lines while circling around them. The result is a helix, like a stretched spring, not a circle. On the exam, check the geometry before assuming r = mv/(qB) describes the whole motion.
A charged particle moving perpendicular to a uniform magnetic field travels in a circle because the magnetic force qvB always points toward the center, acting as the centripetal force.
Setting qvB equal to mv²/r gives the radius formula r = mv/(qB), which is the core equation for every circular motion problem in Unit 12.
The magnetic force does zero work on the particle because it's always perpendicular to the velocity, so the particle's speed and kinetic energy never change.
Doubling the magnetic field strength halves the radius of the circle, while doubling the particle's speed or mass doubles it, so know these proportional relationships cold.
Use the right-hand rule for the force direction, and flip the result for negative charges, since positive and negative particles curve in opposite directions in the same field.
Measuring the radius of a particle's circular path in a known field lets you find its charge-to-mass ratio, which is how mass spectrometers work.
It's the path a charged particle follows when moving perpendicular to a uniform magnetic field. The magnetic force F = qvB acts as the centripetal force, producing a circle of radius r = mv/(qB) at constant speed.
No. The magnetic force is always perpendicular to the particle's velocity, so it does zero work and changes only the direction of motion. The particle's speed and kinetic energy stay constant the entire time.
The radius is cut in half. From r = mv/(qB), radius is inversely proportional to field strength, so a stronger field bends the particle into a tighter circle. This exact proportional-reasoning question is a multiple-choice staple.
In a velocity selector, crossed electric and magnetic fields balance so the particle travels in a straight line at one specific speed. In pure circular motion there's only a magnetic field and no balancing force, so the particle curves. Problems often chain them, selecting a speed first and then measuring the circle's radius.
Set the magnetic force equal to the centripetal force, qvB = mv²/r, and solve for v = qBr/m. Given charge, field, radius, and mass, it's one rearrangement, which is exactly how the multiple-choice stems frame it.
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