An electron moves in a circle of radius 10 cm with a constant speed of \( 4 \times 10^6 \, \text{m/s} \). The electric current at a point on the circle is, given \( e = 1.6 \times 10^{-19} \, \text{C} \)

Electron in Circular Path – Current Calculation

Question: An electron moves in a circle of radius 10 cm with a constant speed of \( 4 \times 10^6 \, \text{m/s} \). The electric current at a point on the circle is, given \( e = 1.6 \times 10^{-19} \, \text{C} \)

a) \( 2 \times 10^{-12} \, \text{A} \)

b) \( 1.019 \times 10^{-12} \, \text{A} \)

c) \( 1 \times 10^{-13} \, \text{A} \)

d) \( 1 \times 10^{-14} \, \text{A} \)

Detailed Solution:

We know that an electron revolving in a circular orbit constitutes an equivalent current given by:

\[ I = \frac{e}{T} \] where \( T \) is the time period of revolution.

The time period is: \[ T = \frac{2\pi r}{v} \] Given: \( r = 10 \, \text{cm} = 0.1 \, \text{m} \), \( v = 4 \times 10^6 \, \text{m/s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \)

Substituting into the formula: \[ T = \frac{2\pi \times 0.1}{4 \times 10^6} = \frac{0.2\pi}{4 \times 10^6} \approx 1.57 \times 10^{-7} \, \text{s} \]

So, \[ I = \frac{1.6 \times 10^{-19}}{1.57 \times 10^{-7}} \approx 1.019 \times 10^{-12} \, \text{A} \]

Correct answer: (b) \( 1.019 \times 10^{-12} \, \text{A} \)