A velocity selector has an electric field of magnitude 2620 N/C,directed vertica

Question

A velocity selector has an electric field of magnitude 2620 N/C,directed vertically upward, and a horizontal magnetic field that isdirected south. Charged particles, traveling east at a speed of6.75 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 +3.30 x10-12 C enters the velocity selector traveling east, thenet force (due to the electric and magnetic fields) acting on it is2.20 x 10-9 N, pointing directly upward. What is thespeed of this particle?

Explanation / Answer

The magnetic force and the electric force have reversedirections. Fm=Bqv and Fe=Eq while Fm and Fe is the magnetic and electric forcerespectively. with v0=6,75e3 m/s Fm=Fe so Bv0=E so B=E/v0=2620/6,75e3=0,388T When the particle have other velocity, the net force is not zeroand have the same direction with the electric force. So Fe>Fm Fe-Fm=F (E-Bv1)q=F so E-Bv1=2,2e-9/3,3e-12=6,66e2(V/m) so v1=(E-666)/B plug in the value of B and E v1=5,04e3(m/s)

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