D3-5. Charged Particles in Electric & Magnetic Fields

1. Crossed Fields & Charge-to-Mass Ratio

Balancing the Fields

$$F_E = F_B \implies qE = qvB \implies v = \frac{E}{B}$$
  • When uniform electric and magnetic fields are arranged perpendicularly ("crossed"), they can exert opposing forces on a charged particle.
  • If field strengths are adjusted perfectly, the forces cancel out. The particle moves in a straight line with constant speed.
  • This configuration functions as a Velocity Selector.

J.J. Thomson Experiment (Finding $q/m$)

  • Thomson used this exact principle to find the charge-to-mass ratio of the electron.
  • First, he balanced the fields to find the speed: $v = \frac{E}{B} = \frac{V}{Bd}$ (since $E = V/d$).
  • Next, he turned off the electric field. The electrons began to curve in a circle due to the magnetic field.
  • Using $q/m = v / (rB)$ and substituting $v$, he derived: $$\frac{q}{m} = \frac{V}{r B^2 d}$$

2. Examples

Example 1

Problem: An electron passes between two parallel metal plates moving with a constant velocity of $2.1 \times 10^7\text{ m s}^{-1}$. The potential difference between the plates is $3100\text{ V}$. A uniform magnetic field of magnitude $0.054\text{ T}$ acts perpendicular to the electric field and the movement of the electron to perfectly balance the forces. Calculate the separation of the plates.

Solution:

Step 1: Because the forces are balanced and velocity is constant, $E = vB$.
$$E = (2.1 \times 10^7\text{ m s}^{-1}) \times (0.054\text{ T}) = 1.134 \times 10^6\text{ N C}^{-1}$$
Step 2: Use the electric field between parallel plates equation $E = V/d$ to solve for $d$.
$$d = \frac{V}{E} = \frac{3100\text{ V}}{1.134 \times 10^6\text{ N C}^{-1}} = \mathbf{2.73 \times 10^{-3}\text{ m}}$$