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 2.74 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2530 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 5.71 V/m, (b) in the negative z direction and has a magnitude of 5.71 V/m, and (c) in the positive x direction and has a magnitude of 5.71 V/m?

Explanation / Answer

  For a particle moving in the +y and in a -x B field the force will be in the + z direction

and magnitude = q*v*B = 1.60x10^-19*2530*2.74x10^-3 = 1.10x10^-18N

a) the electric field creates a force = E*q in the +z direction

Fe = 5.71*1.60x10^-19 = 9.136x10^-19N

So F = 1.10x10^-18 + 9.136x10^-19 = 2.0136x10^-18N

b) Now F = 1.10x10^-18 - 9.136x10^-19 = 1.864x10^-19N

c) Now add them as vectors so

F = sqrt((1.10x10^-18)^2 + (9.136x10^-19)^2) = 1.42x10^-18N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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