When at rest, a proton experiences a net electromagnetic force of magnitude 8.6×

Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.6×1013 N pointing in the positive x direction. When the proton moves with a speed of 1.6×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.3×1013 N , still pointing in the positive xdirection.

Find the magnitude of the electric field. Express your answer using two significant figures.

Find the direction of the electric field.

Find the magnitude of the magnetic field. Express your answer using two significant figures.

Find the direction of the magnetic field.

Explanation / Answer

E = FE / e

= (8.6 x 10-13) / (1.6 x 10-19)

E = 5.38 x 106 N/C

[ in the x direction as E is parallel to F]

----------------------------------------------------

B = (FE – Fnet) / q v

B = [(8.6 x 10-13) – (7.3 x 10-13)] / (1.6 x 10-19 * 1.6 * 106)

B = 0.507 T

[ in the z direction as B is perpendicular to F ]

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