A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.41 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2750 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 4.42 V/m, (b) in the negative z direction and has a magnitude of 4.42 V/m, and (c) in the positive x direction and has a magnitude of 4.42 V/m? ** I Need part c, but I cant put it in terms of i+k; so how do I do that cause I only have one black for the answer

Explanation / Answer

magnetic force= QvB = 1.6X10^-19 X 2750 X 3.41X10^-3 = 1.5X10^-18 N in the positive Z direction . a. if electric field is there, Electric force= EQ= 4.42 X 1.6X10^-19 = 7.02 X 10^-19 N Fnet= Fmag + Felec c. if the electric field is in ppositive X direction, Fnet= = 7.02 X 10^-19 N i + 1.5X10^-18 N k magnitude = 1.165 X10^-18 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.