Determine the energy of the incident electron.

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are r1 and r2. The trajectories are perpendicular to a uniform magnetic field of magnitude B. Determine the energy of the incident electron. (Use any variable or symbol stated above along with the following as necessary: me for the mass of elecron and e for the electron's charge.)

K = 

asked Feb 10, 2012 in Physics by Joey33 ~Expert~ (1,216 points)

1 Answer

+2 votes
Best answer

For each electron, the magnetic force is:

magnetic force = q*v*B*sin(90) = m*v^2 / r  (*where q=charge, v=velocity, B=magnetic field, m=mass, and r=radius)

solving the equation for velocity we get:

v = e*B*r / m  (*replaced q with e, which represents charge of electron)


Now set up equation for the energy of the incident electron.

The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic and hence:

K = (1/2)*m*(v1i)^2 + 0  =   (1/2)*m*(v1f)^2 + (1/2)*m*(v2f)^2  (* v1i = initial velocity of elctron 1, and v2f=final velocity of electron 2...)

plugging in the variables for velocity we get:

K = (1/2)*m*(e*B*r1 / m)^2 +  (1/2)*m*(e*B*r2 / m)^2 =  ( e^2*B^2 / 2m)*(r1^2 + r2^2)


Here is final answer in simplified form:

answered Feb 18, 2012 by kirby ~Expert~ (3,020 points)
edited Mar 8, 2012 by pokemonmaster