A velocity selector has an electric field of magnitude 2487 N/C, directed vertic

Question

A velocity selector has an electric field of magnitude 2487 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.15 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +4.04 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.71 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

For the first part, we can find the B field using the fact that the magnetic force (qvB) must equal the electric force (qE).

Thus qvB = qE (q cancels)

(6.15 X 10^3)(B) = (2487)

B = .40439 T

For the second part, the net force = 1.71 X 10^-9, so...

qE - qvB = 1.71 X 10^-9

(4.04 X 10^-12)(2487) - (4.04 X 10^-12)(v)(.40439) = (1.71 X 10^-9)

v = 5103 m/s (5.10 X 10^3 m/s)

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