An electron in a circular orbit of radius 0.05 mm performs `10^(16) "rev"//s.` the magnetic moment due to this rotation of electron is `(in A-m^(2)).`

Options

A. `2.16xx10^(-23)`
B. `3.21xx10^(-22)`
C. `3.21xx10^(-24)`
D. `1.26xx10^(-23)`

Explanation

To find the magnetic moment due to the rotation of an electron in a circular orbit, we can follow these steps:

### Step 1: Identify the given values
- Radius of the orbit, \( R = 0.05 \, \text{mm} = 0.05 \times 10^{-3} \, \text{m} = 5 \times 10^{-5} \, \text{m} \)
- Frequency of revolution, \( f = 10^{16} \, \text{rev/s} \)
- Charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the area of the circular orbit
The area \( A \) of a circle is given by the formula:
\[ A = \pi R^2 \]
Substituting the value of \( R \):
\[ A = \pi (5 \times 10^{-5})^2 \]
Calculating this:
\[ A = \pi (25 \times 10^{-10}) = 25\pi \times 10^{-10} \approx 7.85 \times 10^{-10} \, \text{m}^2 \]

### Step 3: Calculate the current due to the electron's rotation
The current \( I \) due to the revolving electron can be calculated using the formula:
\[ I = q \cdot f \]
Substituting the values:
\[ I = (1.6 \times 10^{-19} \, \text{C}) \cdot (10^{16} \, \text{rev/s}) = 1.6 \times 10^{-3} \, \text{A} \]

### Step 4: Calculate the magnetic moment
The magnetic moment \( \mu \) is given by the product of the current and the area:
\[ \mu = I \cdot A \]
Substituting the values of \( I \) and \( A \):
\[ \mu = (1.6 \times 10^{-3} \, \text{A}) \cdot (7.85 \times 10^{-10} \, \text{m}^2) \]
Calculating this:
\[ \mu \approx 1.26 \times 10^{-22} \, \text{A m}^2 \]

### Final Answer
The magnetic moment due to the rotation of the electron is approximately:
\[ \mu \approx 1.26 \times 10^{-22} \, \text{A m}^2 \]

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