Question

An electron (charge = –1.6 x 10–19 C) is moving at 3.0 x 105 m/s in the positive x direction. A magnetic field of 0.80 T is in the positive z direction. The magnetic force on the electron is

Answer 1

Answer:

F = 3.84*10^-34N ^j

Explanation:

In order to calculate the magnetic force on the electron you use the following formula:

$\vec{F}=q\vec{v} \ X\ \vec{B}$       (1)

q: charge of the electron = -1.6*10^-19C

B: magnitude of the magnetic field = 0.80T

v: speed of the charge = 3.0*10^5 m/s

The direction of the motion of the electron is in the +i^ direction (+x direction). The magnetic field is in the +^k direction (+z direction).

The cross product between these unit vectors, by taking into account the minus sign of the charge, is given by:

-^i X ^k = ^j

The magnetic force is in the ^j direction (+y direction).

The magnitude of the magnetic force is:

$F=qvB=(1.6*10^{-19}C)(3.0*10^5m/s)(0.80T)=3.84*10^{-34}N$

The magnetic force on the electron has a magnitude of 3.84*10^-34N in the +y direction