Motion of Charged Particles in a Magnetic Field

When charged particles move through a conductor, a magnetic field is produced on that area of the conductor and this magnetic field exerts force on the moving charges. The motion of the charges can still be explained using the Newton’s laws of motion.

Let us consider a positive charge q at point P moving with velocity $$\vec v$$ in a uniform magnetic field which is directed into the plane as shown in the figure below. The figure also shows the direction of the force.

Since $$\vec v$$ and $$\vec B$$ are perpendicular, the magnetic force is given by the equation $$\vec F = q \vec v \times \vec B$$ and its magnitude is $$F = qvB$$.

The magnitude of the velocity of the charged particle is not affected by the magnetic force but the direction is. This means that the magnetic force can never have any component parallel to the velocity and thus, it never does any work to the charged particle. This also applies even if the magnetic field is not uniform.

This principle of the charged particle’s motion can be stated as

A charged particle always moves with constant speed if acted only by the magnetic field.

Looking back at the figure, we see that as the charged particle moves around the magnetic field, only the direction of the force and velocity changes but their magnitudes do not. Their magnitude remains constant since they are always perpendicular to each other. Using the concept of circular motion, in which the speed is constant, the centripetal acceleration is $$v^2/R$$ and applying the Newton’s second law, we have the equation for the magnetic force as

$$F = |q|vB=m{v^2 \over R}$$         (1)

where m is the mass of the charged particle.

Rearranging and simplifying equation (1), we derive the equation for the radius of a circular orbit, R as

$$R = {mv \over |q|B}$$     (2)

If the moving charged particle is negative, then the motion will be clockwise around the orbit.

Now, we combine equation (2) and the equation for speed of a particle in circular motion which is $$v=R\omega$$, to get the equation for the angular speed of the particle in the magnetic field as

$$\omega = {v \over R} = v{|q|B \over mv} = {|q|B \over m}$$             (3)

The number of revolutions per unit time is the frequency, f given by the equation $$f = {\omega \over 2\pi}$$. It is not dependent on the radius of the path.

In a uniform magnetic field, the direction of the direction of the velocity is not perpendicular to the magnetic field. It then follows that the parallel component of the velocity to the field is constant since there is no force which is parallel to the field. In this case, the charged particle is moving in a helical path. See Figure below.

Figure

The radius of the helix can be solved using equation (2) where v is the component of the velocity which perpendicular to the magnetic field.

If the magnetic field is not uniform, the motion of the charged particles would be more complicated. Consider a charged particle moving in a magnetic field produced between two coils as shown in the figure.

Figure

A particle located near any of the coil will experience a magnetic field which is directed towards the center of the region. The particle will then move back and forth of the magnetic field region in a spiral form. Since this set-ep may confine charged particles, it is also called as magnetic bottle.