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 2.74 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2860 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is in the positive z direction and has a magnitude of 4.05 V/m, in the negative z direction and has a magnitude of 4.05 V/m, and in the positive x direction and has a magnitude of 4.05 V/m?

Explanation / Answer

For a particle moving in the +y and in a -x B field the force will be in the + z direction

and magnitude = q*v*B = 1.60x10^-19*2860*2.74x10^-3 = 12.538*10^-19 N

a) the electric field creates a force = E*q in the +z direction

Fe = 4.05*1.60x10^-19 = 6.48x10^-19N

So F = 1.2538x10^-18 + 6.48x10^-19 = 19.018x10^-19 N

b) Now F = 12.538x10^-19 - 5.14x10^-19 = 6.058x10^-19N

c) Now add them as vectors so

F = sqrt((12.538x10^-19)^2 + (6.48x10^-19)^2) = 14.11x10^-19 N

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