Magnetic field strength (B) = 6.5 G = $6.5 \times {10^{ – 4}}T$

Speed of the electron (v) = $4.8 \times {10^6}m/s$

Charge on the electron (e) = $1.6 \times {10^{ – 19}}C$

Mass of the electron (m_e) = $9.1 \times {10^{ – 31}}kg$

Angle formed by the fired electron and the magnetic field, $\theta  = {90^ \circ }$

The magnetic force exerted on the electron in a magnetic field is given by the relationship:$F = evB\sin \theta $

The travelling electron receives centripetal force from this force. As a result, the electron begins to move in a circular route of radius r.

Hence, centripetal force exerted on the electron,

${F_e} = \frac{{m{v^2}}}{r}$

The centripetal force imparted on the electron is equal to the magnetic force at equilibrium, i.e.,

F_e = F

$\frac{{m{v^2}}}{r} = evB\sin \theta $

$r = \frac{{mv}}{{eB\sin \theta }}$

So,

$r = \frac{{9.1 \times {{10}^{ – 31}} \times 4.8 \times {{10}^6}}}{{6.5 \times {{10}^{ – 4}} \times 1.6 \times {{10}^{ – 19}} \times \sin {{90}^ \circ }}}$

$r = 4.2cm$

As a result, the radius of the electron’s circular orbit is 4.2 cm.