Question

X 107 kg) 27.25 A proton (q 1.60 x 1019 C, m 1.67 moves in a uniform magnetic field B (0.500 T)1. At t = 0 the proton has velocity components ux-1.50 × 105 m/s, vy=0, and vz = 2.00 × 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 104 V/m)t. (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?

Answer 1

a) We know that the magnetic force on the charge particle is given by 1.6x10-19 x 1.5 0 2 x10 0.5 0 0 1.6x 10-"jN So there is a force of 1.6x1014 Nacting on the proton in the negative y direction. b) Yes, the proton will have an acceleration in the direction of the applied field 1.6x10-19 x 2 x 10% 1.67 x10-27 a 1.9x10 m/s c) The proton takes a helical path. A circular path due to Magnetic field and it accelerates in the x direction due to the electric field. The magnetic force would make the proton take a circular path with radius 1.67 x10 x2x10 1.6x10-19 x 0.5 r 0.0042 m As the electric field increases the velocity of the proton in the direction of the magnetic field it has no effect on the radius of the helical motion.

d) The time period is V. 2π x 0.0042 2x105 0.13x10-6 s So at half this time the proton will travel some distance. From equations of motion we know at t- we get T aT^ 0.13x10 1.9x10x(0.13x10-9) -0.009750.00401 s0.0138 m The x component of displacement is 0.0138 m.