A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.79 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2890 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.23 V/m, (b) in the negative z direction and has a magnitude of 3.23 V/m, and (c) in the positive x direction and has a magnitude of 3.23 V/m?

Explanation / Answer

a) force = Eq - q vB = q (E-vB) = 7.732 *10^-19 N in the negative direction b) force = Eq + q vB = 1.8 *10^18 N in negative z direction c) force = -q vB k + Eq i = -1.29*10^-18 k + 5.168*10^-19 N force = 1.38*10^-18 N

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