In a chamber, a uniform magnetic field of
Question:

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

Solution:

Magnetic field strength, B = 6.5 G = 6.5 × 10–4 T

Speed of the electron, v = 4.8 × 106 m/s

Charge on the electron, e = 1.6 × 10–19 C

Mass of the electron, me = 9.1 × 10–31 kg

Angle between the shot electron and magnetic field, θ = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB sinθ

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,

$F_{\mathrm{c}}=\frac{m v^{2}}{r}$

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

$F_{\mathrm{c}}=F$

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

$r=\frac{m v}{B e \sin \theta}$

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

$=4.2 \times 10^{-2} \mathrm{~m}=4.2 \mathrm{~cm}$

Hence, the radius of the circular orbit of the electron is 4.2 cm.

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