Question

15. Find the force on an electron crossing a uniform magnetic field of intensity 0.5 T WILII

velocity of 106 ms-1. The charge carried by the electron is -1.6 x 10-19C.​

Answer 1

F = 8.48 x
`10^(-18)` N

Explanation:

The force on an electron in a magnetic field can be determined by;

F = qvB Sin θ

where: e is the electron, v is its velocity, B is the measure of the magnetic field, and θ is the angle of its path.

Thus,

q = -1.6 x
`10^(-19)` C

v = 106 m/s

B = 0.5 T

θ =
`90^(o)`

So that,

F = 1.6 x
`10^(-19)` x 106 x 0.5

= 8.48 x
`10^(-18)` N

The force on the electron crossing the uniform magnetic field is 8.48 x
`10^(-18)` N.