Question

3. In a situation like that shown below, the charged particle is a proton
(q = 1.60 x 10-19 C, m = 1.67 x 10-27 kg) and the uniform, 0.500-T magnetic field is directed along the x-axis. At t =0 the proton has velocity components vx = 1.5 x 105 m/s, vy = 0 and vz = 2.00 x 105 m/s. Only the magnetic force acts on the proton. (a) At t =0, find the force on the proton and its acceleration. (b) Find the radius of the resulting helical path, the angular speed of the proton, and the pitch of the helix (the distance traveled along the helix axis per revolution).

Answer 1

(a) Only z-component of the velocity will cause the magnetic force in the direction of y-axis of magnitude:

$$F=qv_zB=1.6\cdot10^{-14}\text{ N}.$$

According to Newton's second law, the acceleration is

$$a=\frac{F}{m}=9.6\cdot10^{12}\text{ m/s}^2.$$

(b) Under influence of the Lorentz force calculated above the proton begins circular motion in yOz plane with radius derived from the following equation describing the equilibrium between the Lorentz and the centripetal forces:

$$qv_zB=m\frac{v_z^2}{r},$$

$$r=\frac{mv_z}{qB}=4.169\cdot10^{-3}\text{ m}.$$

The angular speed:

$$\omega=\frac{v_z}{r}=4762000\text{ s}^{-1}.$$

The time it takes to make one circle:

$$T=\frac{2\pi}{\omega}=1.319\cdot10^{-7}\text{ s}.$$

Meanwhile (of course, you haven't forgotten that the proton also moves along x-axis) the pitch is:

$$h=v_xT=0.01979\text{ m}.$$