Review Conceptual Example 2 as an aid in understanding this problem. A velocity

Question

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude 2223 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.63 103 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 +3.88 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.85 10-9 N, pointing directly upward. What is the speed of this particle?

Explanation / Answer

let magnitude of magnetic field be B. then when it goes without being deflected, q*E=q*v*B so B=E/v=2223/(6.63*10^3)=0.3352 T so when net force=1.85*10^(-9) N, q*E-q*v*B=1.85*10^(-9) so v=5.209*10^3 m/s

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