An electron is revolving around a proton in a circular orbit of diameter `1 A^(@)`. If it produces a megnetic field of 14 `wb//m^(2)` at the proton, then its angular velocity will be about

Options

A. `4.375 xx 10^(16)` rad/s
B. `10^(10)` rad/s
C. `4 xx 10^(16)` rad/s
D. `10^(15)` rad/s

Answer

To solve the problem, we need to find the angular velocity of an electron revolving around a proton, given that it produces a magnetic field of 14 Wb/m² at the proton. The diameter of the orbit is given as 1 Å (angstrom).

### Step-by-Step Solution:

1. **Determine the Radius of the Orbit**:
The diameter of the orbit is given as 1 Å. To find the radius (R), we divide the diameter by 2.
\[ R = \frac{1 \, \text{Å}}{2} = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \]

2. **Define the Current (I)**:
The electron, as it revolves, creates a current. The current (I) can be defined as the charge (e) flowing through the loop per unit time. The time period (T) for one complete revolution is given by:
\[ T = \frac{2\pi}{\omega} \]
Therefore, the current can be expressed as:
\[ I = \frac{e}{T} = \frac{e \omega}{2\pi} \]

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

4. **Substituting the Current into the Magnetic Field Equation**:
Substitute the expression for current (I) into the magnetic field equation:
\[ B = \frac{\mu_0 \left(\frac{e \omega}{2\pi}\right)}{2R} \]
Rearranging gives:
\[ B = \frac{\mu_0 e \omega}{4\pi R} \]

5. **Rearranging for Angular Velocity (ω)**:
We can rearrange the equation to solve for \(\omega\):
\[ \omega = \frac{4\pi B R}{\mu_0 e} \]

6. **Substituting Known Values**:
We know:
- \(B = 14 \, \text{Wb/m}^2\)
- \(R = 0.5 \times 10^{-10} \, \text{m}\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)

Substituting these values into the equation for \(\omega\):
\[ \omega = \frac{4\pi \times 14 \times (0.5 \times 10^{-10})}{(4\pi \times 10^{-7}) \times (1.6 \times 10^{-19})} \]

7. **Calculating ω**:
Simplifying the above expression:
\[ \omega = \frac{14 \times 0.5 \times 10^{-10}}{10^{-7} \times 1.6 \times 10^{-19}} \]
\[ = \frac{7 \times 10^{-10}}{1.6 \times 10^{-26}} = \frac{7}{1.6} \times 10^{16} \approx 4.375 \times 10^{16} \, \text{rad/s} \]

### Final Answer:
The angular velocity \(\omega\) is approximately \(4.375 \times 10^{16} \, \text{rad/s}\).