Answer to Question #159798 in Electric Circuits for zain ul abdeen

Question

Review problem. One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.40 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.044 0 T. Determine the energy (in keV) of the incident electron.

Answer 1

The velocity of the incident electron after the interaction can be found from the radius of its trajectory:

$v=\frac{eBR}{m}.$

Its kinetic energy is

$E_K=\frac12 mv^2=\frac12 \frac{(eBR)^2}{m}.$

The energy in keV will be

$E_{K\text{(kEV)}}=\frac{E_K}{1000e}=\frac{eB^2R^2}{2000m}=5.51\cdot10^{-22}\text{ keV}.$