In AP Physics 2, the radius of curvature is the radius of the circular path a charged particle follows when it moves perpendicular to a uniform magnetic field, given by r = mv/(qB). It comes from setting the magnetic force qvB equal to the centripetal force mv²/r.
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force F = qvB sin θ (with θ = 90°) acts at a right angle to the velocity at every instant. A force that's always perpendicular to motion doesn't change speed, it changes direction. That's the recipe for uniform circular motion, and the radius of that circle is the radius of curvature.
The equation isn't handed to you, you build it. Set the magnetic force equal to the required centripetal force: qvB = mv²/r. Solve for r and you get r = mv/(qB). Read the proportionalities like a story. More mass or more speed means more inertia fighting the turn, so the circle gets bigger. More charge or a stronger field means a harder magnetic pull, so the circle tightens. This single relationship is why mass spectrometers can sort particles by mass, why isotopes land in different spots, and why a faster particle sweeps a wider arc in the same field.
Radius of curvature 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. It's the payoff of essential knowledge 12.2.B.2, since F = qvB sin θ only becomes useful for predicting motion once you combine it with Newton's second law for circular motion. This is also one of the AP exam's favorite synthesis moments, because it forces you to merge a Unit 12 force with the circular motion framework you learned back in AP Physics 1. If you can derive r = mv/(qB) instead of just memorizing it, you're doing exactly what the FRQ rubrics reward.
Keep studying AP® Physics 2 Unit 12
F_B = qvB sin θ (Unit 12)
This is the force that bends the path. When velocity is perpendicular to the field, sin θ = 1, and qvB becomes the centripetal force. The radius of curvature equation is just this force plugged into F = mv²/r.
Circular motion and centripetal acceleration (Unit 12)
The magnetic force never speeds the particle up or slows it down because it's always perpendicular to velocity. It only turns the particle, which is the exact definition of centripetal force. Same physics as a ball on a string, with the field playing the role of the string.
Charge-to-mass ratio (Unit 12)
Rearrange r = mv/(qB) and the radius depends on m/q. Two particles with the same speed in the same field follow different circles only if their mass-to-charge ratios differ, which is the whole operating principle of a mass spectrometer.
Velocity selector (Unit 12)
A velocity selector uses crossed electric and magnetic fields to let through only particles with one specific speed. Feed those particles into a magnetic field region afterward and v is fixed, so the radius of curvature now sorts purely by mass and charge.
The exam loves comparison setups. The 2024 Short FRQ Q4 gave two particles, one with mass M and charge −Q and another with mass M/2 and charge +2Q, and asked you to compare their motion in the same field. To answer, you derive r = mv/(qB) and reason about ratios, plus use the right-hand rule (remembering negative charges curve the opposite way). Multiple-choice questions pull the same move with isotopes, like C-12 and C-14 ions entering a field with identical velocity. Same charge and speed means the same force, but the heavier isotope has a larger radius. Expect to: derive r from qvB = mv²/r, compare radii using proportional reasoning, and determine curve direction with the right-hand rule. Pure plug-and-chug is rare; ratio logic is the real test.
The phrase shows up twice in AP Physics 2 and means totally different things. In geometric optics, radius of curvature describes how sharply a mirror or lens surface is curved, and it relates to focal length (R = 2f for spherical mirrors). In Unit 12, it describes the circular path of a charged particle in a magnetic field, r = mv/(qB). Context tells you which one a question wants. Charged particles and B-fields mean the magnetism version; mirrors, lenses, and focal points mean the optics version.
The radius of curvature for a charged particle moving perpendicular to a uniform magnetic field is r = mv/(qB).
You derive it by setting the magnetic force equal to the centripetal force, qvB = mv²/r, which is the step FRQ rubrics actually credit.
Larger mass or speed makes the circle bigger, while larger charge or a stronger field makes it tighter.
The magnetic force does no work on the particle, so its speed stays constant and only its direction changes.
Two ions with the same charge and speed but different masses (like C-12 and C-14) feel the same force but travel circles of different radii, which is how mass spectrometers separate isotopes.
Use the right-hand rule for the curve direction, and flip the result for negative charges.
It's the radius of the circular path a charged particle follows when moving perpendicular to a uniform magnetic field, given by r = mv/(qB). It comes from equating the magnetic force qvB with the centripetal force mv²/r.
No. The magnetic force is always perpendicular to the velocity, so it does zero work and the speed never changes. The field only bends the path into a circle (or helix), which is why the particle has a constant radius of curvature.
No, just the same name. In geometric optics, radius of curvature describes the curvature of a mirror or lens surface and connects to focal length. In Unit 12 magnetism, it describes a charged particle's circular path and equals mv/(qB).
If both ions have the same charge and enter with the same speed, they feel the same magnetic force qvB. But C-14 has more mass, so the same force produces less centripetal acceleration and a wider circle, since r = mv/(qB) grows with m.
It's better to know how to derive it. Set F = qvB equal to mv²/r and solve for r. FRQs, like the 2024 short FRQ comparing two particles with different mass and charge, reward showing that derivation and using ratio reasoning rather than quoting the formula.
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