A point charge of 5.0 × 10–7 C moves to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T
Physics
Lilmissbored
Question
A point charge of 5.0 × 10–7 C moves to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T. What is the magnitude of the magnetic force acting on the charge? 0 N 2.3 × 10–3 N 23 N 2.3 × 1011 N
2 Answer

1. User Answers Arnold11
So we want to know what is the magnitude of the magnetic force F on charge Q=5.0*10^7 C if the charge is moving with speed v=2.6*10^5 m/s perpendicular to the magnetic field B=1.8*10^2 T. So the formula for the Lorentz force is F=Q*v*B, where Q is the charge, v is the speed of the charge and B is the magnitude of the magnetic field. If the speed of the charge is parallel with the magnetic field, the force is 0 N. I'm going to assume that the screen and also the magnetic field B is perpendicular to the speed of the charge. So we simply plug in the numbers in the formula and get the magnitude of the force F=0.0023 N = 2.3*10^3 N 
2. User Answers Numinum
Answer:
B
Explanation:
answer is b