Question

An electron moving at 5.06 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)

Answer 1

`8.1^(\circ), 171.9^(\circ)`

Step-by-step explanation:

The magnitude of the magnetic force exerted on the moving electron is:

`F=qvB sin \theta`

where here we have

`F=1.40\cdot 10^(-16) N` is the magnitude of the force

`q=1.6\cdot 10^(-19) C` is the magnitude of the electron charge

B = 1.23 T is the magnetic field intensity

`\theta` is the angle between the direction of the electron's velocity and the magnetic field

Solving the equation for
`\theta`, we find:

`sin \theta = (F)/(qvB)=(1.40\cdot 10^(-16)N)/((1.6\cdot 10^(-19) C)(5.06\cdot 10^3 m/s)(1.23 T))=0.141`

which gives the following two angles:

`\theta = 8.1^(\circ)\\\theta = 180^(\circ)-8.1^(\circ) = 171.9^(\circ)`