PLEASE HELP 200 points for CORRECT answer. THE ANSWER IS ALSO NOT THE NEGATIVE V

Question

PLEASE HELP 200 points for CORRECT answer.

THE ANSWER IS ALSO NOT THE NEGATIVE VERSION OF THIS ANSWER EITHER

Two charged particles are moving with equal velocities of 4.70 m/s in the +x-direction. At one instant of time the first particle with a charge of -4.600 HC is located at x 0 and y +3.90 cm, and the second particle with a charge of -8.70 HC is located at x 0 and y 3.90 cm What is the y-component of the magnetic force on the first particle due to the second? 1.13e-15 N Submit Answer You have entered that answer before Incorrect. Tries 2/7 Previous Tries

Explanation / Answer

magnetic field due to a moving charge q2 = 8.7 uC is B2 = (mu_o/4*pi)*(q2*v/r^2) = 10^-7*(8.7*10^-6*4.7)/(0.078^2) = 6.72*10^-10 T

magnetic force on first particle is F = q1*v*B2*sin(90) = 4.6*10^-6*4.7*6.72*10^-10 = 1.45*10^-14 N along negative y-axis

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