Question

An electron of kinetic energy 45 keV moves in a counter-clockwise circular orbit perpendicular to a magnetic field of 0.325 T. Calculate the acceleration of the electron.

Answer 1

Let's convert the kinetic energy of the electron from keV to Joule:
$K=45 \cdot 10^3 eV \cdot 1.6\cdot 10^{-19} J/eV =7.2 \cdot 10^{-18}J$
And from the kinetic energy, we find the velocity of the electron:
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the electron mass. Re-arranging the formula, we find v:
$v= \sqrt{ \frac{2K}{m} } =3.98 \cdot 10^6 m/s$

The magnetic force acting on the electron is:
F=qvB
where 
$q=1.6 \cdot 10^{-19} C$ is the electron charge
$v=3.98 \cdot 10^6 m/s$ is the electron velocity
B=0.325 T is the magnetic field.

For Newton's second law, we also have that F=ma, so we can rewrite the equation as
$ma=qvB$
and so from this, we can find the acceleration (which is a centripetal acceleration, because the electron is moving by circular motion):
$a= \frac{qvB}{m}=2.27 \cdot 10^{17} m/s^2$