Question

A particle with a charge of-2.50 X 10-8 C is moving with an instantaneous velocity of magnitude 40.0 km/s in the xy- plane at an angle of 50° counterclockwise from the +x axis. What are the magnitude and direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the (a) -x direction, and (b) +z direction?

Answer 1

The expression for magnetic force exerted on the moving particle due to magnetic field is F = q v B sin phi (b) Use the expression of magnitude of magnetic force acting on a moving charged particle to solve the problem. The direction of magnetic field and velocity of the particle shown Convert the unit velocity of particle from km/s to m/s v = 40.0 km/s = 40 km/s (10^3 m/1 km) v = 40.0 times 10^3 m/s \r\n From fig (a) phi = 130 degree F = q v B sin phi sub values F = (2.50 times 10^-8 C) (40 times 10^3 m/s) (2.00 T) sin 130 degree F = 1.53 times 10^-3 n The force on negatively charged particle is the negative z- direction. Hence the force on the negatively charged particle is pointed in the z- direction d) The magnetic field is perpendicular to xy- plane then it pointed to + z direction then phi = 90 degree F = q v B sin phi