Question

A cosmic ray electron moves at 7.45 â 106 m/s perpendicular to the earth's magnetic field at an altitude where field strength is 1.10 â 10â5 t. what is the radius (in m) of the circular path the electron follows?

Answer 1

The magnetic force on the electron is:
$F=qvB \sin \theta$
where q is the electron charge, v the electron speed, B the magnetic field intensity and $\theta$ is the angle between v and B. Since the electron is traveling perpendicular to the magnetic field, $\sin \theta = \sin 90^{\circ}=1$, so we can write 
$F=qvB$

This force acts as centripetal force of the motion, $F_c = m \frac{v^2}{r}$, where m is the electron mass and r the radius of the circular orbit. So we can write:
$qvB=m \frac{v^2}{r}$
and by re-arranging this equation, we find the radius of the circular orbit, r:
$r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31}kg)(7.45 \cdot 10^6 m/s)}{(1.6 \cdot 10^{-19}C)(1.10 \cdot 10^{-5}T)}= 3.85 m$