In hydrogen atom, an electron is revolving in the orbit of radius `0.53 Å` with `6.6xx10^(15) rotations//second`. Magnetic field produced at the centre of the orbit is

Options

A. `0.125 Wb//m^(2)`
B. `1.25 Wb//m^(2)`
C. `12.5 Wb//m^(2)`
D. `125 Wb/m^(2)`

Answer

To find the magnetic field produced at the center of the orbit of an electron revolving in a hydrogen atom, we can use the formula derived from the Biot-Savart law. Here are the steps to solve the problem:

### Step 1: Understand the Parameters
We have the following parameters:
- Radius of the orbit, \( R = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \)
- Frequency of rotation, \( n = 6.6 \times 10^{15} \, \text{rotations/second} \)
- Charge of the electron, \( Q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Use the Formula for Magnetic Field
The magnetic field \( B \) at the center of a circular loop due to a current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 3: Calculate the Current \( I \)
The current \( I \) can be expressed in terms of charge and frequency:
\[ I = n \cdot Q \]
Substituting the values:
\[ I = (6.6 \times 10^{15}) \cdot (1.6 \times 10^{-19}) \, \text{A} \]

### Step 4: Substitute Values into the Magnetic Field Formula
Now, substitute \( I \) into the magnetic field formula:
\[ B = \frac{\mu_0 (n \cdot Q)}{2R} \]
Substituting the known values:
\[ B = \frac{4\pi \times 10^{-7} \cdot (6.6 \times 10^{15} \cdot 1.6 \times 10^{-19})}{2 \cdot (0.53 \times 10^{-10})} \]

### Step 5: Calculate the Magnetic Field
Now, we need to calculate the values step by step:
1. Calculate \( n \cdot Q \):
\[ n \cdot Q = 6.6 \times 10^{15} \cdot 1.6 \times 10^{-19} = 1.056 \times 10^{-3} \, \text{A} \]
2. Substitute this into the magnetic field equation:
\[ B = \frac{4\pi \times 10^{-7} \cdot 1.056 \times 10^{-3}}{2 \cdot (0.53 \times 10^{-10})} \]
3. Calculate the denominator:
\[ 2 \cdot (0.53 \times 10^{-10}) = 1.06 \times 10^{-10} \]
4. Now calculate \( B \):
\[ B = \frac{4\pi \times 10^{-7} \cdot 1.056 \times 10^{-3}}{1.06 \times 10^{-10}} \]
5. Finally, calculate the value:
\[ B \approx 12.5 \, \text{T} \, (\text{Tesla}) \]

### Final Answer
The magnetic field produced at the center of the orbit is approximately:
\[ B \approx 12.5 \, \text{T} \]

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