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

Answer:

Approximately $3.7 \times 10^{-4}\; {\rm T}$.

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 $F = 5.9 \times 10^{-18}\; {\rm N}$ and $v = 9.9 \times 10^{4}\; {\rm m\cdot s^{-1}}$. The magnitude of the charge on an electron is $q = 1.602 \times 10^{-19}\; {\rm C}$ (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 $B$:

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

(All values are measured in standard units.)