At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction. When viewed from a point on the positive z axis, it subsequent motion is:

A proton moves perpendicularly to a uniform magnetic field at 3.0 ✕ 107 m/s and exhibits an acceleration of 1.0 ✕ 1013 m/s2 in the +x-direction when its velocity is in the +z-direction. Determine the magnitude and direction of the field.

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.12 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2960 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 3.55 V/m, (b) in the negative z direction and has a magnitude of 3.55 V/m, and (c) in the positive x direction and has a magnitude of 3.55 V/m? (a) Number i (b) Number i (c) Number i Units Units Units >

A proton moves from rest in the +z direction after being accelerated by the potential difference V. The proton then passes undeflected through a region with a uniform electric field E in the +x direction and a uniform magnetic field B in the +y direction. If the experiment is repeated using a potential difference of 2V so that the proton is accelerated more, how would the proton move? (A) deviates in the y direction. B) It deviates in the +y direction. c) It deviates in the +x direction. D It deviates in the-x direction.