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: v x =2.00*10^ 6 m / s ,v y =0 and v z =3.00*10^ 6 m / s . Find the radius and pitch of the helical path followed by the electron.

Answer 1

Given:

$B=0.500\:\rm T$

$v_x=2.00*10^6\:{\rm m/s},\; v_y=0,\;v_z=3.00*10^6\:{\rm m/s}$

The radius of helical path

$$R=\frac{mv_z}{eB}=\frac{9.1*10^{-31}*3.00*10^6}{1.6*10^{-19}*0.500}=3.4*10^{-5}\:\rm m$$

The period of rotation

$$T=\frac{2\pi m}{eB}=\frac{6.28*9.1*10^{-31}}{1.6*10^{-19}*0.500}=7.14*10^{-11}\:\rm s$$

The pitch of the helical path

$$h=v_xT=2.00*10^6*7.14^{-11}=1.43*10^{-4}\:\rm m$$