Question: A proton (q = 1.60 X 10-19 C, m = 1.67 X 10-27 kg) moves in a uniform magnetic field B = (0.500 T) i.

A proton (q = 1.60 X 10-19 C, m = 1.67 X 10-27 kg) moves in a uniform magnetic field B = (0.500 T) i. At t = 0 the proton has velocity components Ux = 1.50 X 105 m/s, Uy = 0, and v. = 2.00 X 105 m/s (see Example 27.4).
(a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +x-direction, E = (+2.00 X 104v/m)i.
(b) Will the proton have a component of acceleration in the direction of the electric field?
(c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain.
(d) At t = T/2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0?

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a IDENTIFY and SET UP Eq274 gives the total force on the proton At t0 F q B q vi vk Bi qvBj F 160109 ... View full answer

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