Question

An electron moves with a speed of 5.0 x 10^4m/s

perpendicularto a uniform magnetic field of .20T. What is the
magnitude ofthe magnetic force on the electron?

Answer 1

`F=1.6* 10^(-15)\ N`

Step-by-step explanation:

Given that,

Speed of the electron,
`v=5* 10^4\ m/s`

Magnetic field, B = 0.20 T

We need to find the magnitude of magnetic force on the electron. The formula for the magnetic force is given by :

`F=qvB\ sin\theta`

Here,
`\theta=90^(\circ)`

`F=qvB`, q is the charge on electron

`F=1.6* 10^(-19)* 5* 10^4* 0.2`

`F=1.6* 10^(-15)\ N`

So, the magnitude of the magnetic force on the electron is
`1.6* 10^(-15)\ N`. Hence, this is the required solution.