A proton is occupying a region where there is an electric field (E=195 N/C) in t

Question

A proton is occupying a region where there is an electric field (E=195 N/C) in the positive y-direction and a magnetic field (B=11.7 mT) directed out of the page. The directions and strengths of these fields do not change, but the forces created by one or both of them may change. The picture above shows the path taken by the proton due to these influences assuming it started from the origin (point A). a.) What is the magnitude and direction of the total force on the proton at point A (where v=30 km/s)? It is standard to state the direction as an angle measured from the positive x-axis. b.) What is the work done on the proton in moving it from point A to point B? Assume the vertical scale is .0513m. c.) What is the magnitude and direction of the total force on the proton at point B? d.) What is the work done on the proton in moving it from point A to point C? e.) Indicated the directions of the electric and magnetic forces at point D?

Explanation / Answer

None of these answers are correct. JollySalt, I'm actually in the same class as you...I'm assuming. Part A - The work is not "cumulative." You need to use the pythagrium theorum...so find the force of the E-field and the force of the B-field, square them, then add them....and then take the square root of that number. Part B - You need to use the work energy theorum (change in KE=Work), so that you have the final velocity (which is the velocity at point B...which you need to find the forces on the proton at point B) Part C - You're going to find out when you use the right-hand-rule that the force of the B-field is straight down and the force of the E-field is straight up. So you just subtract the B-field force from the E-field force, and the direction 270 degrees from the positive x-axis. Part D - Here there is no work done Part E - use the right-hand-rule. Fe is still straight up, and Fb is down and left.

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