A proton of mass m, and charge q is in a box that contains an electric field E, and the box is located in Earth's magnetic field BEarth. The proton moves with an initial velocity v vertically upward from the surface of Earth. Assume gravity is negligible.
a. Indicate the direction of the electric field inside the box so that there is no change in the trajectory of the proton while it moves upward in the box. Explain your reasoning.
b. Determine the speed of the proton while in the box if it continues to move vertically upward. Express your answer in terms of the fields and the given quantities.
The proton now exits the box through the opening at the top.
c. Sketch the path of the proton after it leaves the box.
d. Determine the magnitude of the acceleration a of the proton just after it leaves the box, in terms of the given quantities and fundamental constants.
Right hand rule
F = qvBsin[tex]\theta[/tex]
The Attempt at a Solution
a) Direction of a field is usually determined using the right hand rule, correct?
b) Could I solve for v in F = qvBsin[tex]\theta[/tex] ?
c) This would depend on (a)
Any help? I would like some guidance as to how to approach this problem.