A force that steers but never speeds up

The right-hand rule computes $\vec{v}\times\vec{B}$, which is a purely geometric object. It knows nothing about the sign of the charge. The sign enters afterward, in the multiplication:

$$\vec{F} = q\,\vec{v}\times\vec{B}.$$

So the procedure is fixed:

1. Point the fingers along $\vec{v}$, the ACTUAL velocity of the particle, not the conventional current direction.
2. Curl toward $\vec{B}$. The thumb gives $\vec{v}\times\vec{B}$.
3. If $q > 0$, that is the force. If $q < 0$, reverse it. Once.

The word "once" is doing real work. The most common electron-beam error is a double flip: reversing the velocity because "electron flow is opposite to current" and then reversing again for the negative charge, which lands back on the positive-charge answer with two errors that hid each other. Pick one bookkeeping and audit it: if you drew the electron's real velocity, you flip exactly one time, at the end.

The left-hand rule is an equivalent shortcut for negative charge, and it is safe as long as you never mix conventions inside one problem. What is not safe is switching hands halfway because an answer looked wrong.

Work is $dW = \vec{F}\cdot d\vec{r} = \vec{F}\cdot\vec{v}\,dt$. For the magnetic force,

$$\vec{F}\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0,$$

because $\vec{v}\times\vec{B}$ is perpendicular to $\vec{v}$ by construction. This is not an approximation and it does not depend on the field being uniform or steady in space. A magnetic field cannot change $|\vec{v}|$, cannot change kinetic energy, and cannot accelerate a particle in the everyday sense of "make it faster".

Use this as a diagnostic. If a problem has a free charge speeding up or slowing down, something other than $q\vec{v}\times\vec{B}$ did it: an applied electric field, an induced electric field from a changing $\vec{B}$ (Unit 13), or a collision. A "magnetic brake" slows a moving conductor, but the chain there runs through induced currents and the electric fields that drive them, never through a magnetic force doing work on a free charge.

With $|\vec{v}|$ fixed and $\vec{F}$ always perpendicular to it and of constant magnitude $|q|v_\perp B$, the perpendicular part of the motion is uniform circular motion. Any velocity component along $\vec{B}$ feels no force and simply persists, so the general path is a helix: a circle in the perpendicular plane, drifting at constant $v_\parallel$ along the field.

Set the magnetic force equal to the centripetal requirement for the perpendicular motion:

$$|q|v_\perp B = \frac{mv_\perp^2}{r} \quad\Longrightarrow\quad r = \frac{mv_\perp}{|q|B}.$$

Now get the period by dividing the circumference by the speed, and watch what cancels:

$$T = \frac{2\pi r}{v_\perp} = \frac{2\pi m}{|q|B}, \qquad \omega = \frac{|q|B}{m}.$$

The speed is gone. A faster particle runs a proportionally bigger circle and arrives back at the same moment, so every particle of a given $m$ and $q$ in a given field orbits with the same period, no matter how fast it is going. That is the cyclotron principle: a fixed-frequency alternating voltage across the dees stays in step with the particle through every one of its ever-widening turns.

Two habits keep the scaling straight:

1. Write $r = mv_\perp/(|q|B)$ and read off which symbol is in the numerator before answering a "what happens if" question. $v$ upstairs means faster gives a BIGGER circle.
2. For a mass spectrometer, hold constant what the apparatus holds constant. At fixed speed, $r \propto m/|q|$. At fixed accelerating voltage instead, $v$ itself depends on $m$, and $r \propto \sqrt{m/|q|}$. Those are different answers to different experiments.

Put $\vec{E}$ and $\vec{B}$ perpendicular to each other and send a charge through perpendicular to both. The two forces are

$$F_E = |q|E \quad \text{(independent of speed)}, \qquad F_B = |q|vB \quad \text{(grows with speed)}.$$

Arrange the geometry so they oppose. They balance at exactly one speed:

$$|q|E = |q|vB \quad\Longrightarrow\quad v = \frac{E}{B}.$$

Everything about this device follows from noticing which force depends on $v$ and which does not:

1. A particle entering FASTER than $E/B$ has the magnetic force winning, and deflects to the side the magnetic force points.
2. A particle entering SLOWER has the electric force winning, and deflects the other way.
3. The charge $|q|$ cancels, so the selected speed is the same for every particle. So does the mass, which never appears: a heavy ion and an electron at $v = E/B$ both pass straight through.
4. Reversing the sign of the charge reverses BOTH forces together, so the balance survives. The selector picks a speed, not a species.

Feed the emerging beam into a region of pure $\vec{B}$ and $r = mv/(|q|B)$ with $v$ now known turns a landing position into a mass. That two-stage arrangement is the mass spectrometer.