Question

An electron moves north at a velocity of 9.9 x 10^4 m/s and has a magnetic force of 5.9 x 10^-18 N west exerted on it. If the magnetic field points upward, what is the magnitude of the magnetic field?

Answer 1

Approximately
`3.7 * 10^(-4)\; {\rm T}`.

Step-by-step explanation:

When an electric point charge moves through a magnetic field, magnitude of the magnetic force on the charge would be:

`F = q\, v\, B\, \cos(\theta)`, where:

• 
  `q` is the magnitude of the charge;
• 
  `v` is the magnitude of velocity of the charge relative to the magnetic field;
• 
  `B` is the magnitude of the magnetic field;
• 
  `\theta` is the angle between velocity
  `v` and magnetic field
  `B`.

In this question, it is given that
`F = 5.9 * 10^(-18)\; {\rm N}` and
`v = 9.9 * 10^(4)\; {\rm m\cdot s^(-1)}`. The magnitude of the charge on an electron is
`q = 1.602 * 10^(-19)\; {\rm C}` (also known as the elementary charge.)

Since the velocity of the electron (north) is perpendicular to the magnetic field (upwards,) the angle between the two would be
`\theta = 90^(\circ)`.

Rearrange the equation
`F = q\, v\, B\, \cos(\theta)` to find the magnitude of the magnetic field
`B`:

`\begin{aligned}B &= (F)/(q\, v\, \cos(\theta)) \\ &= (5.9 * 10^(-18))/((1.602 * 10^(-19))\, (9.9 * 10^(4))\, \cos(90^(\circ))) \\ &\approx 3.72 * 10^(-4)\; {\rm T}\end{aligned}`.

(All values are measured in standard units.)