Question

In a chamber, a uniform magnetic field of $6.5\, G \,(1 G = 10 ^{–4} T)$ is maintained. An electron is shot into the field with a speed of $4.8 × 10^6 m s^{–1 }$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $(e = 1.5 × 10^{–19} C, m_e = 9.1×10^{–31} kg)$

Answer 1

Magnetic field strength, $B = 6.5\,G = 6.5 × 10^{–4}\, T$
Speed of the electron, $v = 4.8 × 10^6 ms^{-1}$
Charge on the electron, $e = 1.6 × 10^{–19} C$
Mass of the electron, $m_e = 9.1 × 10^{–31} kg$
Angle between the shot electron and magnetic field, $θ = 90\degree$
Magnetic force exerted on the electron in the magnetic field is given as:
$F = evBsinθ$
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
$F_e = \frac{mv^2}{r}$
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
$F_e = F$
$\frac{mv^2}{r} = evBsinθ$
$r = \frac{mv}{eBsinθ}$
So,
$r =\frac {9.1 × 10^{-31} × 4.8 × 10^6 }{6.5 × 10^{-4} × 1.6 × 10^{-19} × sin 90\degree}) = 4.2 × 10^{-2}\,m = 4.2\,cm$
Hence, the radius of the circular orbit of the electron is 4.2 cm.