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.