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 2.70 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1950 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 4.55 V/m, (b) in the negative z direction and has a magnitude of 4.55 V/m, and (c) in the positive x direction and has a magnitude of 4.55 V/m?

Explanation / Answer

Magnetic force acting on a charge (q) particle is given as F = q(VxB) where V and B are the velocity of particle and magnetic field strength. x is the cross product. Magnetic force F = 1.6*10^-19 *( 1950*2.7*10^-3) = 8.424*10^-19 N and directed in the positive z direction a)Electric force = 4.55*1.6*10^-19 = 7.28*10^-19 since electric field is directed in positive direction so net force is 7.28*10^-19 + 8.424*10^-19 N = 15.704 b)Electric force = 4.55*1.6*10^-19 = 7.28*10^-19 since electric field is directed in positive direction so net force is - 7.28*10^-19 + 8.424*10^-19 N = 1.144 c) Electric force = 4.55*1.6*10^-19 = 7.28*10^-19 Net force = 10^-19 *sqrt(7.28^2 + 8.424^2) = 11.133830248 * 10^-19 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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