Charge in a Uniform Magnetic Field

Suppose you have a positive charge moving in a uniform magnetic field, perpendicular to the field. The following things are true:
A charge in a uniform magnetic field will undergo uniform circular motion.
• 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.
In short, a charge moving in a uniform magnetic field will undergo uniform circular motion.

Uniform Circular Motion

Because the motion and the magnetic field are proportional to each other, the force on the charge is $$\abs{\vec F}=\abs{q\vec v\times \vec B}=qvB$$ and the charge's acceleration is $$a={F\over m}={q\over m}vB$$ Recall from mechanics that the centripetal acceleration in uniform circular motion is related to the radius of the circle and the speed of the object: $$a={v^2\over r}$$ Put those together and we get $${q\over m}vB={v^2\over r}$$ We can solve this to find the radius of the circle that the charge makes in the magnetic field:
$$r=\frac{mv}{\abs{q}B}$$
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.
A negatively charged particle enters a uniform magnetic field that is perpendicular to the screen. It takes the solid path (path B) shown.
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.)

Period and Frequency

A charge in a uniform magnetic field will spin around in a circle forever at the same speed $$v$$. During each trip around, it travels a distance $$d=2\pi r$$, and so the time it takes to go around once is $$T={d\over v}=\frc{v}(2\pi r)=\frc{v}2\pi {mv\over qB}\implies$$
$$T=2\pi {m\over qB}$$
Identical charges in an identical field will orbit with the same frequency
This is the period of the charge's orbit. We can also find the frequency of its orbit:
$$f=\frc{T}=\frc{2\pi}{\abs{q}B\over m}$$
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. $$\omega=2\pi f={\abs{q}B\over m}$$