A force that only turns

$$F = qvB\sin\theta, \qquad \theta \ \text{between} \ \vec{v} \ \text{and} \ \vec{B}.$$

Two cases give exactly zero, and both are worth checking before anything else:

1. The charge is at rest. $v = 0$, so $F = 0$, in the strongest field you can build.
2. The charge moves along the field lines. $\theta = 0$, $\sin 0 = 0$, so $F = 0$ however fast it goes.

Only the component of $\vec{v}$ perpendicular to $\vec{B}$ produces a force. That is a real difference from the electric case, where $\vec{F} = q\vec{E}$ acts whether or not the charge moves, and it is why one field can steer a beam hard while another leaves it alone.

The symmetry runs the other way too: a stationary charge produces no magnetic field of its own. Magnetism is what electricity looks like when charge is moving.

Set up the right-hand rule with the fingers along $\vec{v}$, curling toward $\vec{B}$; the thumb gives the force on a positive charge. The result is perpendicular to the plane that $\vec{v}$ and $\vec{B}$ share.

So an answer with the force arrow parallel to $\vec{B}$ is wrong by construction, and so is one parallel to $\vec{v}$. Both mistakes produce a particle that speeds up or reverses, and the sideways deflection that actually happens never appears.

Then the step that is easiest to lose: for a negative charge, reverse the result. The hand motion feels like the whole procedure, so the sign gets dropped. Say it out loud as its own step. The check is that a proton and an electron sent along the same path through the same field bend to opposite sides.

A force perpendicular to the velocity does no work, so the magnetic force on a moving charge changes the direction of $\vec{v}$ and never its magnitude. A particle enters a field region at $2\times10^6$ m/s and leaves at exactly $2\times10^6$ m/s.

A stronger field tightens the curve and still adds no energy, which is why a particle can circle in a uniform field indefinitely at one speed. Watching the velocity vector turn and calling that "acceleration" in the everyday sense is what produces the wrong answer; acceleration here means a change of direction.

So if a problem's kinetic energy does change, the energy came from somewhere else: an electric field, or an induced emf. The magnetic force on a moving charge can never supply it.

For a charge moving perpendicular to a uniform field, the magnetic force supplies exactly the centripetal force:

$$qvB = \frac{mv^2}{r} \qquad \Longrightarrow \qquad r = \frac{mv}{qB}.$$

Read every factor off the solved expression rather than arguing from the force. Twice the speed needs twice the radius; twice the mass, twice the radius; twice the charge or twice the field, half the radius.

"A faster particle feels a bigger force, so it gets held in tighter" is the trap, and the resolution is in the two powers of $v$. The magnetic force grows as $v$, and the centripetal force required grows as $v^2$. The requirement outruns the supply, so the circle has to open up.

One tidy consequence: the period $T = 2\pi m/(qB)$ contains no $v$ at all. Faster particles travel bigger circles in exactly the same time.