Answer to Question #312733 in Physics for cols

Question

At the instant an electron enters a uniform magnetic field of 0.500 T directed along the positive x-axis, its velocity has the following components; vx 2.00 x 10⁶m/s, vy = 0, and Vz = 3.00 x 10⁶ m/s. Find the radius and pitch of the helical path followed by the electron

Answer 1

The radius of path

$$r=\frac{mv_z}{qB_x}=\frac{9.1*10^{-31}*3.0*10^6}{1.6*10^{-19}*0.50}=3.4*10^{-5}\:\rm m$$

The period of rotation

$$T=\frac{2\pi r}{v_z}=\frac{2\pi*3.4*10^{-5}}{3.0*10^6}=7.1*10^{-11}\:\rm s$$

Hence, the pitch of helical path

$$h=v_xT\\ =2.0*10^6*7.1*10^{-11}=1.4*10^{-4}\:\rm m$$