A velocity selector has an electric field of magnitude 2310 N/C, directed vertic

Question

A velocity selector has an electric field of magnitude 2310 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 7.45 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 +5.00 x 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 2.06 x 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

Electric field E = 2310 N/C
Speed v = 7.45 x 10 3 m/s The particle is not deflected. Force on particle due to electric field = Force on particle due to magnetic field                      Eq = Bvq From this magnetic field B = E / v                                         = 0.31 T (b).Electric charge q ' = +5.00 x 10 -12 C The net force (due to the electric and magnetic fields) acting on it F = 2.06 x 10 -9 N The speed of this particle v ' = ? We know F = Eq - Bvq     Since both the forces are opposite to each other                     = (2310 x +5.00 x 10 -12 ) - (0.31 x v x +5.00 x 10 -12 )                     = 11500 x 10 -12  - ( 1.5503 x 10 -12 x v )                     = 10 -12 ( 11500 - 1.5503 x v ) 2.06 x 10 -9 = 10 -12 ( 11500 - 1.5503 x v )             2060 = ( 11500 - 1.5503 x v )       1.5503 v = 9490                   v = 6121.39 m / s                      

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