Chapter 28, Problem 010 A proton travels through uniform magnetic and electric f

Question

Chapter 28, Problem 010 A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.95 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2200 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 4.09 V/m, (b) in the negative z direction and has a magnitude of 4.09 v/m, and (c) in the positive x direction and has a magnitude of 4.09 V/m? (a) Number (b) Number (c) Number Units Units Units

Explanation / Answer

Magnetic force acting on a charge (q) particle, F = q (VxB)

Magnetic force, F = 1.6 x 10-19 ( 2200 x 2.95 x 10-3) = 1.04 x 10-18 N (directed in the positive z direction)

a) Electric force, E = 4.09 x 1.6 x 10-19 = 6.54 x 10-19

since electric field is directed in positive direction so net force is (6.54 x 10-19) + (1.04 x 10-18) = 1.69 x 10-18 N

b) Electric force, E = 4.09 x 1.6 x 10-19 = 6.54 x 10-19

since electric field is directed in positive direction so net force is (-6.54 x 10-19) + (1.04 x 10-18) = 3.86 x 10-19 N

c) Electric force = 4.09 x 1.6 x 10-19 = 6.54 x 10-19

Net force = 10-19 Sqrt(6.542 + 10.42) = 12.285 x 10-19 N

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