A velocity selector has an electric field of magnitude 2670 N/C, directed vertic

Question

A velocity selector has an electric field of magnitude 2670 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.00 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 +4.10 x 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.65 x 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

The electric field E = 2670 N/C The speed of the charged particle is v = 6000 m/s Let B be the applied magnetic field Then, The net force acting on the particle due to both the fields is F = q(E+Bv)               (B and v are perpendicular to each other) As the charged particle are undeflected, the net force on the particle will be zero Therefore E = -Bv B = -E/v B = -2670 N/C / 6000 m/s B = -0.445 T Now the particles with charge q = 4.1*10-12 C entered the velocity selector The net force on the particles is F = 1.65*10-9 N Therefore F = q(E+Bv) 1.65*10-9 N = (4.1*10-12 C)[2670 N/C + (-0.445 T)(v)] v = 5095.64 m/s

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