Chapter 28 Problem 010 A proton travels through uniform magn

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 1.96 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1640 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric s e) in the positve zdectnnd has a magnitude of 3.67 V/m, (b) in the negative has a magnitude of 3.67 V/m? (a) Number (b) Number (c) Number Show Work is REQUIRED for this question: Units Units i Units Open Show Work

Solution

a) f = q(E+v*B)

F = 1.6*10^-19(3.67k+1.96*10^-3*1640k) = 1.1*10^-18 k N

b) f = 1.6*10^-19(-3.67k+1.96*10^-3*1640k) =-0.72*10^-19 k N

c) f = 1.6*10^-19(3.67i+1.96*10^-3*1640k) = (5.87i+5.14k)*10^-19 N