Answer:
Approximately [tex]3.7 \times 10^{-4}\; {\rm T}[/tex].
Explanation:
When an electric point charge moves through a magnetic field, magnitude of the magnetic force on the charge would be:
[tex]F = q\, v\, B\, \cos(\theta)[/tex], where:
In this question, it is given that [tex]F = 5.9 \times 10^{-18}\; {\rm N}[/tex] and [tex]v = 9.9 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex]. The magnitude of the charge on an electron is [tex]q = 1.602 \times 10^{-19}\; {\rm C}[/tex] (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 [tex]\theta = 90^{\circ}[/tex].
Rearrange the equation [tex]F = q\, v\, B\, \cos(\theta)[/tex] to find the magnitude of the magnetic field [tex]B[/tex]:
[tex]\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}[/tex].
(All values are measured in standard units.)
