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

Question #159798

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.


1
Expert's answer
2021-02-03T16:15:46-0500

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


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

Its kinetic energy is


EK=12mv2=12(eBR)2m.E_K=\frac12 mv^2=\frac12 \frac{(eBR)^2}{m}.

The energy in keV will be


EK(kEV)=EK1000e=eB2R22000m=5.511022 keV.E_{K\text{(kEV)}}=\frac{E_K}{1000e}=\frac{eB^2R^2}{2000m}=5.51\cdot10^{-22}\text{ keV}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment