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 6.33 103 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 1.92 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

First find the magnitude of the magnetic field Since the first particle travels undeflected then E = v*B so B = E/v = 2487/6.33x10^3 = 0.393T The upward force on the second particle = q*E = 4.04x10^-12*2487 = 10.05x10^-9N If the net force is 10.05x10^-9 upward then the downward force from the magnetic field = 10.05x10^-9 - 1.92x10^-9 = 8.13x10^-9N So FB = q*v*B = 8.13x10^-9N so v = 8.13x10^-9/(4.04x10^-12*0.392) = 5133.6m/s

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