When at rest, a proton experiences a net electromagnetic force of magnitude 8.5

Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.5 times 10^-13 N pointing in the positive x direction. When the proton moves with a speed of 1.6 times 10^6 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.4 times 10^-13 N, still pointing in the positive x direction. Find the magnitude of the electric field. Part B Find the direction of the electric field. positive x direction negative x direction positive y direction negative y direction positive z direction negative z direction Part C Find the magnitude of the magnetic field.

Explanation / Answer

a)F = E*q
so E = F/q = (8.5x10^-13 X + 0Y +0Z)N/1.60x10^-19C = 5.31x10^6X + 0Y + 0Z N/C

b) The force due to the electric field does not change due to a movement of the proton but the force due magnetic field does. The F due to the B field is 7.4x10^-13-8.5x10^-13 = -1.1x10^-13N (in the -x direction)

To create a force in the -x direction for a proton moving in the y direction the magnetic field must point in the -z direction

So Bx = 0 and By = 0

Now use F = q*v*Bz

so Bz = F/(q*v) = 11x10^-14/(1.60x10^-19*1.6x10^6) = 0.43T

So B = 0 X + 0 Y - 0.43 Z T

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