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AP Physics C: Electricity and Magnetism · Unit 12 · Topic 1 of 4 · 6 min lesson

Magnetic Force on Charges

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.

The big idea

A magnetic field pushes on a moving charge with F = qv × B. The force is always perpendicular to the velocity, so it changes direction, not speed: a charge moving across a uniform field goes in a circle of radius r = mv/(qB).

1. The force and the circle

r + v F B out of page r = mv/(qB) T = 2πm/(qB)
For a positive charge, point fingers along v, curl toward B (or use the cross product): the force is toward the center. A negative charge curves the other way.
F = qv × B|F| = qvB sin θ
r = mv/(qB)from qvB = mv²/r
T = 2πm/(qB)independent of speed
v = E/Bvelocity selector: qE = qvB
WB = 0F ⟂ v always

2. What the motion looks like

Faster charges make bigger circles but take the same time per lap: r ∝ v, so the circumference grows as fast as the speed. That’s why a cyclotron works at one fixed frequency.

Velocity at an angle to B: the component along B is unaffected, so the path is a helix.

Mass spectrometer: accelerate ions through ΔV (qΔV = ½mv²), then bend them in B. Heavier ions make wider circles: r = (1/B)√(2mΔV/q).

Crossed E and B: only charges with v = E/B pass straight through, whatever their charge or mass.

3. Example, then try it

Example

A proton (m = 1.67 × 10⁻²⁷ kg) moves at 2.0 × 10⁶ m/s perpendicular to a 0.50 T field. Find the radius of its path and the time for one lap.

r = mv/(qB) = (1.67 × 10⁻²⁷)(2.0 × 10⁶) ÷ [(1.6 × 10⁻¹⁹)(0.50)] ≈ 4.2 cm.

T = 2πm/(qB) = 2π(1.67 × 10⁻²⁷) ÷ (8.0 × 10⁻²⁰) ≈ 1.3 × 10⁻⁷ s.

Try it · work it on paper first

In a mass spectrometer, singly charged carbon-12 ions (m = 2.0 × 10⁻²⁶ kg) start from rest, are accelerated through 1000 V, then enter a 0.40 T field at right angles. Find their speed, the radius of their path, and where they land relative to the entry point.

1

Speed: qΔV = ½mv² → v = √(2qΔV/m) = √[2(1.6 × 10⁻¹⁹)(1000) ÷ (2.0 × 10⁻²⁶)] ≈ 1.3 × 10⁵ m/s.

2

Radius: r = mv/(qB) = (2.0 × 10⁻²⁶)(1.26 × 10⁵) ÷ [(1.6 × 10⁻¹⁹)(0.40)] ≈ 3.9 cm.

3

Landing spot: after a half circle, one diameter away: about 7.9 cm. A heavier isotope lands farther out.

Answer: about 1.3 × 10⁵ m/s; r ≈ 3.9 cm; it lands about 7.9 cm from the entrance.

4. Common mistakes

Using the right hand for an electron.Find the direction for +q, then reverse it.
Thinking the magnetic force speeds charges up.It does no work; only the direction changes.
Expecting faster charges to circle more often.The period doesn’t depend on v.
Check yourself

A charged particle circles in a uniform magnetic field. The field strength is doubled (same speed). What happens to the radius?

r = mv/(qB), so doubling B halves r.

Ready to practice?

6 questions on Magnetic Force on Charges, each with a hint and an explanation for every wrong answer.

Unit 12 topics

Every topic has a short lesson or quick review and its own quiz.