In a chamber,a uniform magnetic field of $6.5 \;G \left(1 \;G = 10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $\left(e = 1.6 \times 10^{-19} \;C, m_{e} = 9.1 \times 10^{-31} \;kg \right)$

(N/A) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$. Since the electron enters the field normal to it,the force is always perpendicular to the velocity,acting as a centripetal force. This causes the electron to move in a circular path.
Given:
$B = 6.5 \;G = 6.5 \times 10^{-4} \;T$
$v = 4.8 \times 10^{6} \;m s^{-1}$
$e = 1.6 \times 10^{-19} \;C$
$m_{e} = 9.1 \times 10^{-31} \;kg$
$\theta = 90^{\circ}$
Equating the magnetic force to the centripetal force:
$evB = \frac{m_{e}v^{2}}{r}$
$r = \frac{m_{e}v}{eB}$
Substituting the values:
$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$
$r = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}}$
$r = 4.2 \times 10^{-2} \;m = 4.2 \;cm$
The radius of the circular orbit is $4.2 \;cm$.