How Does a Charged Particle Move in a Uniform Magnetic Field?

For both SL and HL, a charged particle moving perpendicular to a uniform magnetic field follows a circular path at constant speed. If its velocity has a component parallel to the field, it follows a helical path instead.

The Reasoning and Mechanism

A moving charge experiences a magnetic force with magnitude

$$F = |q|vB\sin\theta$$

where $q$ is the charge, $v$ is its speed, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the field.

The force is always perpendicular to both the magnetic field and the particle's velocity. Therefore, it changes the direction of the velocity but not its magnitude. The magnetic field does no work, so the particle's kinetic energy and speed remain constant.

For perpendicular motion, the magnetic force acts as the centripetal force:

$$|q|vB=\frac{mv^2}{r}$$

Therefore,

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

A particle with greater momentum has a larger radius, while a stronger magnetic field produces a smaller radius. The period is

$$T=\frac{2\pi m}{|q|B}$$

The direction is found using the appropriate hand rule for a positive charge. A negative charge curves in the opposite direction.

A common misconception is that the magnetic force speeds up the particle. It does not: because the force is perpendicular to the displacement, it changes direction only.

IB Exam Technique

In Topic D.3, exam questions often ask you to determine the direction of curvature or calculate the radius. State that magnetic force provides the centripetal force, write both equations, equate them, and use the magnitude (|q|) when calculating the radius.