A velocity selector has an electric field of magnitude 2100 N/C, directed vertic

Question

A velocity selector has an electric field of magnitude 2100 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.96 x 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 +3.80 x 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.94 x 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

When the particles are able to go through the velocity selector without being deflected, the magnetic and electric forces balance. Therefore

Fe = Fb

qE = qvB

or v = E/B

Thus we can solve for B

B = E/v

B = 2100/6.96 X 103

B = .302 T

Now, using that value of B, we can answer the next protion. Since the force is directed upward, the electric field is greater than the magnetic field. Therefore

Fnet = Fe - Fb

1.94 X 10-9 = (3.8 X 10-12)(2100) - (3.8 X 10-12)(v)(.302)

Solve for v

v = 5263 m/s

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