- The force is perpendicular to the charge's motion, because \(\vec F=q\vec v\times\vec B\).
- That means that no work is done on the charge.
- No work done on the charge means that the charge's kinetic energy remains constant. (Unlike the electric field, the magnetic field has no associated potential energy it can use to speed up or slow down charges.)
- … and if the kinetic energy \(K=\frc2mv^2\) stays constant, that means that the charge moves at a constant speed.
- Because the force is perpendicular to the motion, the acceleration is perpendicular to the motion as well. This is centripetal acceleration.

$$r=\frac{mv}{\abs{q}B}$$

Notice that the radius gets bigger when the charge has larger
momentum: either the charge is more massive (more inertia) or is
moving faster. The radius gets smaller when you increase the
magnetic field, or if you increase the size of the charge.
The magnetic field points

into the page
out of the page

For the charge to move in the circle shown, the force on the charge
has to point towards the center. At the point where the charge
enters the field, its velocity points to the right, and the force
must point up. Now, if the field pointed into the page, then \(\vec
v\times\vec B\) would point upward; however, because this is a
negative charge, \(q\vec v\times\vec B\) points downward.
The magnetic field must point out of the page.

Suppose the particle had a larger mass. Which path would it have
taken?

A
B
C

The radius of the particle is proportional to the mass: larger
mass, larger radius. (Like a truck, the particle has to take
wider turns.)

$$T=2\pi {m\over qB}$$

$$f=\frc{T}=\frc{2\pi}{\abs{q}B\over m}$$

which is the number of times it spins around per second. (This is
sometimes called the cyclotron frequency). Notice that
it is independent of speed: two charges with the same mass and
charge, in the same magnetic field, will always orbit with the same
frequency (as shown in the figure). If the charge goes around \(f\)
times per second, then it covers \(2\pi f\) radians per second,
and that is the charge's angular velocity, a term
you'll hopefully remember from rotational kinematics.
$$\omega=2\pi f={\abs{q}B\over m}$$