Charged particles in a magnetic field

​​In a nutshell

When charged particles enter a magnetic field perpendicularly, they feel a force which makes their path curve. A velocity selector selects charged particles for a specific velocity by using both an electric and magnetic field.

Equations

Constants

Variable definitions

Force on a charged particle

When a particle beam moves perpendicularly through a magnetic field it feels a force. The force is also perpendicular to the velocity of the particle, and therefore the particle has a curved path.

The force can be calculated using the following equation:

​$$F = BIL$$​​

By substituting in the following:

​$$L = vt \\ I = \dfrac{Q}{t}$$​​

The following equation can also be obtained for the force acting on a single particle:

$$F=BQv$$​​

If the particles in the beam are electrons or protons this can be turned into:

​$$F=Bev$$​​

Where $$e$$ is the elementary charge.​

Note: These equations are for the force acting on a single particle. You would need to multiple the equations by $$N$$ for $$N$$ particles. You can also calculate the acceleration of a particle by using Newton's second law, $$F = ma$$.

Example

A proton enters a magnetic field with a magnetic flux density of $$0.25\ T$$. What is the force acting on the proton if it has a speed of $$6\times10^{4}\ m\ s^{-1}$$?

Firstly, write down what you know:

$$B=0.25\ T\newline v=6\times10^{4}\ m\ s^{-1}$$​​

Next, write down the equation for the force acting on a particle in a magnetic field:

$$F=BQv$$​​

Since the particle is a proton:

$$F=Bev$$​​

Substitute the values into the equation and calculate the force:

$$F=0.25\times1.6\times10^{-19}\times6\times10^{4}=2.4\times10^{-15}\ N$$​​

The proton felt a force of $$\underline{2.4\times10^{-15}\ N}$$ acting on it.

Going in circles

Since the path of the particles is circular, they must feel a centripetal force:

​$$F=\dfrac{mv^2}{r}$$​​

The magnetic force provides the centripetal force on the particle which means that they can be equated. Doing so allows you to find an expression for the radius of the path taken by the particles:

​$$\dfrac{mv^2}{r}=BQv$$​​

​$$r=\dfrac{mv}{BQ}$$​​

Velocity selector

A velocity selector is a device that selects charged particles for a specific velocity by using an electric and a magnetic field. It looks like the following:

1. Vacuum chamber 2. Path of particles with speed $$v$$​​ 3. Particle 4. Magnetic force 5. Electric force 6. Straight direction of motion

​

The plates produce an electric field with field strength $$E$$ which is perpendicular to a magnetic field with flux density $$B$$. When particles enter the chamber, they are acted upon by both of these fields and will deflect in different directions. This deflection will cancel, but only for particles with a specific velocity and will keep a straight path.

The reason why these particles aren't deflected is, due to their velocity, the magnetic force acting on them is equal to the electric force.

​$$F_B=F_E$$​​

You can then substitute these forces and rearrange to find the selected velocities:

​$$BQv=EQ\newline v=\dfrac{E}{B}$$​​

Thus, the velocity of the selected particles can be regulated by changing the electric and magnetic fields.