Question

A uniform 4.4 t magnetic field points north. if an electron moves vertically downward (toward the ground) with a speed of 2.5 × 107 m/s through this field. the charge on a proton is 1.60×10−19 .
a.what is the magnitude of the force acting on it? answer in units of n.

Answer 1

Answer:

$1.8\cdot 10^{-11} N$

Explanation:

The force acting on the electron due to the magnetic field is

$F=qvB sin \theta$

where

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=2.5\cdot 10^7 m/s$ is its speed

$B=4.4 T$ is the intensity of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of the field and the velocity of the electron

Substituting all the numbers into the equation, we find

$F=(1.6\cdot 10^{-19}C)(2.5\cdot 10^7 m/s)(4.4 T) sin 90^{\circ}=1.8\cdot 10^{-11} N$