In an orbit of radius `0.5 A^(0)` an electron revolves with a frequency of `6.25 xx 10^(15)` Hz. The magnetic induction field at its centre is

Options

A. 4`pi` tesla
B. `2pi` tesla
C. 4 tesla
D. 2 tesla

Answer

To solve the problem, we need to find the magnetic induction field (B) at the center of the orbit of an electron revolving in a circular path. We can follow these steps:

### Step 1: Convert the radius from angstroms to meters.
Given:
- Radius \( r = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} = 5 \times 10^{-11} \, \text{m} \)

### Step 2: Identify the frequency of revolution.
Given:
- Frequency \( f = 6.25 \times 10^{15} \, \text{Hz} \)

### Step 3: Calculate the current (I) due to the revolving electron.
The current can be calculated using the formula:
\[ I = \text{Charge} \times \text{Frequency} \]
The charge of an electron \( e = 1.6 \times 10^{-19} \, \text{C} \).
Thus,
\[ I = e \cdot f = (1.6 \times 10^{-19} \, \text{C}) \times (6.25 \times 10^{15} \, \text{Hz}) \]
Calculating this gives:
\[ I = 1.0 \times 10^{-3} \, \text{A} \]

### Step 4: Use the formula for magnetic induction (B) at the center of the circular path.
The formula for the magnetic field at the center of a circular loop carrying current is given by:
\[ B = \frac{\mu_0 I}{2r} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( r = 5 \times 10^{-11} \, \text{m} \)

Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 1.0 \times 10^{-3}}{2 \times 5 \times 10^{-11}} \]

### Step 5: Calculate the magnetic induction (B).
Calculating the above expression:
\[ B = \frac{4\pi \times 10^{-10}}{10 \times 10^{-11}} = \frac{4\pi}{10} \, \text{T} \]
\[ B = 0.4\pi \, \text{T} \approx 1.2566 \, \text{T} \]

### Final Answer:
The magnetic induction field at the center of the orbit is approximately \( 1.2566 \, \text{T} \).

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