Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.50 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0580 T. Determine the energy (in keV) of the incident electron.
__________ keV

Answer 1

After the collision, both of them move perpendicular to the B field in circular paths with the given radii.

Hence, the magnitude of centripetal force will be equal to the magnetic force in the magnetic field.

mvf^2/r = evf1B

=> vf = reB/m

By conservation of energy (in case of elastic collsion)

(1/2)mvi^2 + 0 = (1/2)mvf1^2 + (1/2)mvf2^2

=> KEi = (1/2)m(r1eB/m)^2 + (1/2)m(r2eB/m)^2

=> = e^2B^2/2m(r1^2 + r2^2) = (1.6*10^-19)^2*(0.0580)^2/2*9.1*10^-31[0.01^2 + 0.025^2) = 3.43*10^-14 J

This is equal to 214.083 keV