Solved:
Answered by AI, Verified by Human Experts
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.66 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2930 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.51 V/m, (b) in the negative z direction and has a magnitude of 5.51 V/m, and (c) in the positive x direction and has a magnitude of 5.51 V/m?
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.66 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2930 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.51 V/m, (b) in the negative z direction and has a magnitude of 5.51 V/m, and (c) in the positive x direction and has a magnitude of 5.51 V/m?
Answer:a)b)c)Explanation:We have:B: is the magnetic field = (2.66x10⁻³ T)(-)v: is the velocity of the proton = (2930 m/s)a) To find the magnitude of the net force (F) we need to use the Lorentz force equation:(1)E: is the electric field = (5.51 V/m)q: is the proton charge = 1.6x10⁻¹⁹ CHence, the magnitude of the net force is:b) When E = (5.51 V/m)(), the net force is (equation 1):Then, the magnitude of the net force is:c) When E = (5.51 V/m)(), the net force is:Hence, the magnitude of the net force is:I hope it helps you!...
Unlock full access for 72 hours, watch your grades skyrocket.
For just $0.99 cents, get access to the powerful quizwhiz chrome extension that automatically solves your homework using AI. Subscription renews at $5.99/week.