Ncert Exercises & Solutions

4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

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 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_{e} = \frac{{mv}^{2}}{r}

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

F_{e} = F

\Rightarrow \frac{{mv}^{2}}{r} = evB sinθ

\Rightarrow r = \frac{mv}{eB sinθ}

So,

r = \frac{9.1 \times 10^{- 31} \times 4.8 \times 10^{6}}{6.5 \times 10^{- 4} \times 1.6 \times 10^{- 18} \times \sin 90 °} = 4.2 \times 10^{- 2} m = 4.2 cm

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

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