Example 8.2: Charged particle in magnetic field

Question: Suppose that an electron is accelerated from rest through a voltage difference of $V=10^3$volts and then passes into a region containing a uniform magnetic field of magnitude $B=1.2$T. The electron subsequently executes a closed circular orbit in the plane perpendicular to the field. What is the radius of this orbit? What is the angular frequency of gyration of the electron?
 
Answer: If an electron of mass $m_e =9.11\times 10^{-31}$kg and charge $e=1.60\times 10^{-19}$C is accelerated from rest through a potential difference $V$ then its final kinetic energy is

$$\frac{1}{2}\,m_e\,v^2 = e\,V.$$

Thus, the final velocity $v$ of the electron is given by

$$v = \sqrt{\frac{2\,e\,V}{m_e}} = \sqrt{\frac{(2)\,(1.6\times... ...{(9.11\times 10^{-31})}} = 1.87\times 10^{7}\,{\rm m\,s}^{-1}.$$

The initial direction of motion of the electron is at right-angles to the direction of the magnetic field, otherwise the orbit of the electron would be a spiral instead of a closed circle. Thus, we can use Eq. (167) to calculate the radius $\rho$ of the orbit. We obtain

$$\rho = \frac{m_e\,v}{e\,B} = \frac{ (9.11\times 10^{-31})\,(... ...} {(1.6\times 10^{-19})\,(1.2)} = 8.87\times 10^{-5}\,{\rm m}.$$

The angular frequency of gyration $\omega$ of the electron comes from Eq. (168):

$$\omega = \frac{e\,B}{m_e} = \frac{ (1.6\times 10^{-19})\,(1.... ....11\times 10^{-31})} = 2.11\times 10^{11}\,{\rm rad.\,s}^{-1}.$$