A proton moves eastward (+x) through a region where a uniform magnetic field points northward (+y). Using the Lorentz force law, the magnetic force on the proton points in which direction?
ANorthward — the force aligns with the field
BEastward — the force aligns with the velocity
CUpward (+z) — perpendicular to both velocity and field by the right-hand rule
DZero — the proton must move parallel to the field for a force to act
The magnetic force is F = q(v × B). With v in the +x direction and B in the +y direction, v × B = x̂ × ŷ = ẑ (upward, +z). Since q is positive for a proton, the force is in the +z direction — upward. The force is always perpendicular to both the velocity and the field; it never aligns with either. The common wrong answer (zero) applies only when v is parallel to B, making the cross product zero — which is not the case here.
Question 2 Multiple Choice
Magnetic fields are used to steer charged particle beams in accelerators. Which statement correctly describes what the magnetic field does to the particles?
AIt does no work; the particles' speed stays constant but their direction changes
BIt does positive work, continuously increasing the particles' kinetic energy
CIt does negative work, slowing the particles as they curve
DIt does work proportional to the field strength B
The magnetic force is always perpendicular to the velocity, so the dot product F · v = 0 at every instant. Work requires a force component along the displacement, and the magnetic force has none. This means a magnetic field can only change the direction of motion, never the speed. In a uniform field, this produces circular (or helical) motion with constant speed — the field continuously redirects without adding or removing kinetic energy. Accelerating particles (increasing their speed) requires electric fields, not magnetic fields.
Question 3 True / False
A charged particle moving parallel to a magnetic field experiences zero magnetic force.
TTrue
FFalse
Answer: True
The magnetic force is F = q(v × B). The cross product v × B has magnitude |v||B|sinθ, where θ is the angle between the velocity and field vectors. When v is parallel to B, θ = 0°, so sin(0°) = 0, and the force is zero. Only the component of velocity perpendicular to B contributes to the magnetic force. This is why charged particles in a uniform field travel in helices when they have a velocity component along the field: the parallel component is unaffected (no force), while the perpendicular component is deflected into a circle.
Question 4 True / False
A magnetic field can accelerate a charged particle, increasing its kinetic energy over time.
TTrue
FFalse
Answer: False
This is the key misconception about magnetic forces. The magnetic force F = q(v × B) is always perpendicular to the velocity. Work is defined as W = F · ds, and since the force is perpendicular to the displacement (ds = v dt), the dot product is zero. No work is done, so kinetic energy cannot change. Speed stays constant. The magnetic field is a 'steering' force only — it can curve a particle's path but cannot speed it up or slow it down. Electric fields, not magnetic fields, do work on charges and change their kinetic energy.
Question 5 Short Answer
Why do charged particles move in circular paths in a uniform magnetic field? Explain in terms of what the magnetic force does and does not do to the particle's velocity.
Think about your answer, then reveal below.
Model answer: The magnetic force is always perpendicular to the velocity, so it changes the direction of motion but never the speed (it does no work). This is exactly the condition that produces circular motion: a constant-magnitude force perpendicular to the velocity acts as centripetal acceleration, continuously redirecting the particle without changing how fast it moves. Setting the magnetic force equal to the centripetal force gives qvB = mv²/r, from which the cyclotron radius r = mv/(qB) follows directly.
Circular motion requires a centripetal force — always directed toward the center, always perpendicular to the velocity. The magnetic force satisfies this automatically: it is always perpendicular to v (by the cross product geometry) and has constant magnitude when v and B are perpendicular and B is uniform (since |F| = qvB, and v is constant because no work is done). The radius of the circle encodes particle mass and charge — which is why cyclotrons and mass spectrometers can separate particles by their charge-to-mass ratio.