The magnetic force does no work on a free charge: it turns the velocity and never changes its magnitude

01The mistake

A proton enters a uniform magnetic field perpendicular to its velocity and follows a circular path. Ask whether its speed changes. A large share of a class says it increases, or that it varies around the circle. The speed is constant for as long as the particle is in the field, and the magnetic force does exactly zero work on it.

The energy version is the costlier one on a free-response item. Students write the work done by the magnetic force as $F \cdot d$ using the arc length, getting a large nonzero number. The dot product in $W = \int \vec{F} \cdot d\vec{s}$ is zero at every point because the two vectors are perpendicular at every point.

The cyclotron question exposes it completely. If magnetic forces did work, the magnetic field alone would accelerate the particles and no electric gap would be needed. Students who cannot say what the gap is for have the misconception even if they answer the speed question correctly, because they have not connected the two.

A fourth version appears in the opposite direction and is worth distinguishing: students conclude that since the force does no work, the magnetic field does nothing at all. It changes the direction of the momentum continuously, which is a substantial effect — it is the entire reason the path is a circle.

02Why it makes sense to the student

A force with no work is unprecedented for students at this point. Every force they have computed work for has had a component along the motion somewhere in the problem. Gravity on a projectile does work, friction does work, tension usually does. A force that never does any is a new category.

Circular motion in mechanics was taught with the centripetal force in focus and the zero work rarely mentioned. Students have accepted that uniform circular motion has constant speed with a net inward force, so they already own the idea — it just was not labeled as a statement about work when they learned it.

Strong force reads as big effect, and big effect reads as energy change. A magnetic field that bends a particle into a tight circle is visibly doing something dramatic, and students do not have a category for dramatic-but-energy-neutral.

And the dot product is easy to carry as notation without reading it. $W = \int \vec{F} \cdot d\vec{s}$ gets copied, the vectors are replaced by magnitudes, and the angle silently becomes zero. The perpendicularity is the whole content of the integral and it is the part that disappears first.

03The correction

Chain the two facts and make students say both. The magnetic force is perpendicular to the velocity. A perpendicular force does no work. Therefore the kinetic energy and the speed are constant. Three statements, each following from the one before, and the conclusion is not memorized.

Write the dot product out with the angle in it: $\vec{F} \cdot d\vec{s} = F\,ds\cos 90^\circ = 0$. Making the $\cos 90^\circ$ explicit is what stops the magnitudes-only substitution, and it generalizes to every perpendicular force the student will meet.

Connect it back to the centripetal case they already accept. A ball on a string in uniform circular motion keeps its speed, and the tension does no work for exactly the same reason. Students usually find this persuasive immediately, because they are not being asked to believe something new — only to recognize something they already hold.

Then use the cyclotron as the application that makes the fact load-bearing. The magnetic field supplies the turning and the electric gap supplies the energy, and the design exists because the two jobs cannot be done by the same field. A student who can explain the gap has the concept.

Close the loop on what the field does change. Direction, and therefore the momentum vector, continuously. The radius $r = mv/qB$ follows from Newton's second law with the magnetic force as the centripetal one, and that is a real and useful consequence of a force that does no work.

04A sample question

05What each wrong answer reveals

• A Force over a distance read as work. The dominant wrong answer, and the justification names the reasoning: a force acted through a displacement. Write the dot product with its angle. $\cos 90^\circ = 0$ at every point of the path, so the integral is zero no matter how long the path is.
• B Correct. The magnetic force stays perpendicular to the velocity throughout, so it does no work and the speed is unchanged on exit.
• C Speed and momentum conflated. This student has the work result right, which is the main idea, and has extended it to a vector quantity. The momentum's magnitude is unchanged and its direction rotated by 90 degrees, so the momentum changed substantially. Worth drawing both momentum vectors; the change is as large as the original. A good student making a vector-versus-scalar slip.
• D The dot product evaluated as a product. The same error as A stated as a formula instead of a conclusion, and more diagnostic, since it shows exactly where the angle was dropped. Have the student write the $\cos\theta$ factor and identify $\theta$ at any point on the arc.

A and D are one misconception in two registers, and D is the more useful to see because it localizes the omission. C is a different and much smaller error: the work conclusion is correct and a scalar result was applied to a vector. Treating C like A would be a mis-diagnosis.

06Try it in Mistake Master