Initially at rest,a small copper sphere with a mass of 2.0 * 10^-6 kg and a charge of 3.0 *10^-4 C is accelerated through a 5000 V potential difference before entering a region with a 4.0 T magnetic field that is directed perpendicular to the its velocity.What is the radius of curvature of the copper sphere's motion in the magnetic field region?

Question

Quizplus · Accepted Answer

The correct answer is A because the radius of curvature of the copper sphere's motion in the magnetic field region is given by the equation r = mv/qB, where m is the mass of the sphere, v is its velocity, q is its charge, and B is the magnetic field strength. Plugging in the given values, we get r = (2.0 * 10^-6 kg)(5000 V)/(3.0 * 10^-4 C)(4.0 T) = 2.0 m.

Initially at Rest,a Small Copper Sphere with a Mass of 2.0