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 2.34 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2440 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 3.48 V/m, (b) in the negative z direction and has a magnitude of 3.48 V/m, and (c) in the positive x direction and has a magnitude of 3.48 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*2440*2.34x10^-3 = 9.135x10^-19N

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

Fe = 3.48*1.60x10^-19 = 5.568x10^-19N

So F = 9.135x10^-19 + 5.568x10^-19 = 1.470x10^-18N

b) Now F = 9.135x10^-19 - 5.568x10^-19 = 3.567x10^-19N

c) Now add them as vectors so

F = sqrt((9.135x10^-19)^2 + (5.568x10^-19)^2) = 1.06x10^-18N

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