Question

An electron travels at a speed of 16,748.76 m/s through a uniform magnetic field whose magnitude is 0.0177 T. What is the magnitude of the magnetic force on the electron if its velocity vector and the magnetic field vector are make an angle of 59.24° ?

Answer 1

Magnetic force,
`F=4.07* 10^(-17)\ N`

Step-by-step explanation:

It is given that,

Speed of electron, v = 16748.76 m/s

Magnetic field, B = 0.0177 T

Angle between velocity vector and the magnetic field vector are make an angle of 59.24°. Magnetic force is given by :

`F=qvB\ sin\theta`

`F=1.6* 10^(-19)* 16748.76* 0.0177 \ sin(59.24)`

`F=4.07* 10^(-17)\ N`

So, the magnetic force on the electron is
`4.07* 10^(-17)\ N`. Hence, this is the required solution.