Question

An electron (charge = −1.6ᴇ−19 C) is moving at 3 ᴇ5 m/s in the positive x direction. A

magnetic field of 0.8T is in the positive z direction. The magnetic force on the electron is:
a.0
b.4 ᴇ −14 N, in the positive z direction
c.4 ᴇ −14 N, in the negative z direction
d.4 ᴇ −14 N, in the positive y direction
e.4 ᴇ −14 N, in the negative y direction

Answer 1

The magnetic force on the electron is -3.84 x 10^-14 N in the negative z direction.

Step-by-step explanation:

The magnetic force on an electron moving in a magnetic field can be determined using the equation:
F = q * v * B * sin(theta)

Where:
F is the magnetic force
q is the charge of the electron
v is the velocity of the electron
B is the magnetic field
theta is the angle between the velocity vector and the magnetic field vector.

In this case, the electron has a negative charge
`(q = -1.6 x 10^-19 C),` a velocity of 3 x 10^5 m/s in the positive x direction, and the magnetic field is 0.8 T in the positive z direction.

The angle between the velocity and the magnetic field is 90 degrees, so sin(theta) = 1.

Plugging in the values:

`F = (-1.6 x 10^-19 C) * (3 x 10^5 m/s) * (0.8 T) * (1) = -3.84 x 10^-14 N`

Therefore, the magnetic force on the electron is
`-3.84 x 10^-14 N` in the negative z direction. The correct answer is c.