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 2.48 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2610 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 4.33 V/m, (b) in the negative z direction and has a magnitude of 4.33 V/m, and (c) in the positive x direction and has a magnitude of 4.33 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*2610*2.48x10^-3 = 1.03x10^-8N

a)

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

Fe = 4.33*1.60x10^-19 = 6.928x10^-19N

So

F = 1.03x10^-18 + 6.92x10^-19 = 1.03x10^-18N

b)

Now F = 1.03x10^-18 - 6.92x10^-19 = 7.12x10^-19N

c) Now add them as vectors so

F = sqrt((1.03x10^-18)^2 + (6.92x10^-19)^2) = 1.24x10^-18N

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