Question

An electron is moving with a velocity of 4.3 x 106 m/s in a uniform magnetic field of strength 3.5 T. The velocity vector makes a 55o angle with the magnetic field. What is the magnitude of the magnetic force on the electron?Group of answer choices4.0 x 10-12 N8.0 x 10-12 N2.0 x 10-12 N6.0 x 10-12 N

Answer 1

Given:

The velocity of the electron is

`v=4.3*10^6\text{ m/s}`

The strength of the uniform magnetic field is B = 3.5 T

The angle between the velocity of the electron and the magnetic field is 55 degrees.

Also, the charge on the electron is

`e\text{ =1.6}*10^{-19\text{ }}C`

To find the magnetic force on the electron.

Step-by-step explanation:

The magnetic force can be calculated as

`\begin{gathered} F=\text{Bev sin }\theta \\ =3.5*1.6*10^(-19)*4.3*10^6sin(55^(\circ)\text{)} \\ =2*10^(-12)\text{ N} \end{gathered}`

Final Answer: The magnetic force on the electron is 2 x 10^(-12) N