A velocity selector has an electric field of magnitude 2470 N/C, directed vertic

Question

A velocity selector has an electric field of magnitude 2470 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.50 × 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 +4.00 × 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.90 × 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

Given that the electric field is E=2470 N/C velocity v=6.50 x 103 m/s then the magnetic field in the velocity selector is B=E/v    =0.38T charge of the particle is q=4 x 10-12 C deflecting force F=1.90 x 10-9N if u is the velocity of the particle then F=quB u=F/qB    =1.90 x 10-9/4 x 10-12 x 0.38     =1.25  x 103 m/s

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