An electron accelerates from rest through an electric field into a magnetic field. The plates have a potential

Starting from rest, the electron will have a kinetic energy of 25 electron volts (eV) after passing through the electric field. One eV = 1.602*10^−19 J, and the kinetic energy (neglecting relativistic effects, which are not important here) is 0.5*m*s^2, where m is the rest mass of the electron, and v is its speed.

We have, therefore:

25 eV * 1.602*10^−19 J/eV = 0.5*9.11*10^-31 kg * s^2

s^2 = 8.79*10^12 (m/s)^2

s = 2.96*10^6 m/s

The force on a particle with charge q, moving with vector velocity V in a magnetic field B is given by:

F = qV x B

where "x" means the vector cross product.

If, as it seems in this case, the velocity vector and magnetic field vector are perpendicular to one another, then the magnitude of the force is simply given by the ordinary algebraic product of the charge, the magnitude of the velocity (i.e., the speed), and the magnitude of the magnetic field, so:

|F| = q*|V|*|B|

|F| = 1.60*10^-19 C * 2.96*10^6 m/s * 0.5 T

|F| = 2.37*10^-13 N

(It helps to know that 1 tesla = 1 kg/(sec*coulomb) here).

The force vector is oriented perpendicular to both the velocity and the magnetic field. However, there are still two possible directions that satisfy this condition. The ambiguity can be resolved by using the "right hand rule": point your index finger along the direction of the velocity vector, and your middle finger along the direction of the magnetic field. Your thumb will then point in the direction of the force vector. You'll have to figure out how this applies to what I assume was the figure provided with your problem.

To determine the radius of the electron's orbit, we equate the magnetic force with the centripital force:

q*|V|*|B| = (m*|v|^2)/r

r = (m*|v|^2)/(q*|V|*|B|)

r = (m*|v|)/(q*|B|)

r = (9.11*10^-31 kg * 2.96*10^6 m/s)/(1.60*10^-19 C * 0.5 T)

r = 3.38*10^-5 m

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