Question

Monday Homework Problem 8.9 A pith ball, charged to 10nC, is shot into one of the Pasco lab magnets with speed 20 m such that the velocity is perpendicular to the magnetic field. The field has a strength of 0.25 T. Compute the magnitude of the maximum magnetic force on the pith ball. This may be a very small number

Answer 1

The magnitude of the maximum magnetic force on a pith ball with a charge of 10nC, moving at 20 m/s perpendicular to a 0.25 T magnetic field, is 5×10^-8 N.

Step-by-step explanation:

To compute the magnitude of the maximum magnetic force on a pith ball, one can use the equation F = qvBsin(θ), where F is the force, q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since it is given that the velocity is perpendicular to the magnetic field, θ = 90 degrees and therefore sin(θ) = sin(90) = 1. Plugging in the values given, we have F = 10nC × 20m/s × 0.25T. Since 1nC = 1×10-9 C, the force F = (10×10-9 C) × 20m/s × 0.25T = 5×10-8 N.