A proton is occupying a region where there is an electric field (E = 150 N/C) in

Question

A proton is occupying a region where there is an electric field (E = 150 N/C) in the postive y-direction and a magetic field (B = 0.0090 T) directed out of the page. The picture seen by following this link http://i49.tinypic.com/wt93ll.png shows the path taken by the proton due to these influences assuming it started from the origin (point A). What is the magnitude and direction of the total force on the proton at a point B (where velocity = 40,000 m/s)? State the direction as an angle measured from the positive x-axis.

Explanation / Answer

at point B

net force = qvB - qE = (40000 x 0.0090 - 150) x 1.6 x 10^-19 = 336 x 10^-19 N

magnitude of net force = 3.36 x 10^-17 N

and direction of net force will be along negative y direction

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