A velocity selector has an electric field of magnitude 2500 N/C,directed vertica

Question

A velocity selector has an electric field of magnitude 2500 N/C,directed vertically upward, and a horizontal magnetic field that isdirected south. Charged particles, traveling east at a speed of7.27 x 103 m/s, enter the velocity selector and are ableto pass completely through without being deflected. When adifferent particle with an electric charge of +4.40 x10-12 C enters the velocity selector traveling east, thenet force (due to the electric and magnetic fields) acting on it is2.17 x 10-9 N, pointing directly upward. What is thespeed of this particle?

Explanation / Answer

For 1st particle, net force is zero so we have. Eq=Bqv. so E=Bv. so B=E/v=0,344(T). ------------------------ for 2nd particle, net force is 2,17e-9(N) upward so . Eq-Bqv=2,17e-9. so E-Bv=2,17e-9/4,4e-12=493. so v=(E-493)/B=5,8e3(m/s)

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