An electron is revolving round a proton, producing a magnetic field of `16 weber//m^(2)` in a circular orbit of radius `1 Å. Its angular velocity will be

Options

A. `10^(17)rad//sec`
B. `1//2pixx10^(12)rad//sec`
C. `2pixx10^(12)rad//sec`
D. `4pixx10^(12)rad//sec`

Explanation

To solve the problem of finding the angular velocity of an electron revolving around a proton, producing a magnetic field of \(16 \, \text{weber/m}^2\) in a circular orbit of radius \(1 \, \text{Å}\), we can follow these steps:

### Step 1: Understand the relationship between current and angular velocity
The current \(I\) produced by the revolving electron can be expressed as:
\[ I = \frac{q}{T} \]
where \(q\) is the charge of the electron and \(T\) is the time period of revolution. The time period \(T\) can be related to angular velocity \(\omega\) as:
\[ T = \frac{2\pi}{\omega} \]
Thus, we can rewrite the current as:
\[ I = \frac{q \cdot \omega}{2\pi} \]

### Step 2: Use the formula for the magnetic field produced by a current
The magnetic field \(B\) at the center of a circular loop carrying current \(I\) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where \(\mu_0\) is the permeability of free space and \(r\) is the radius of the orbit.

### Step 3: Substitute for current in the magnetic field equation
Substituting the expression for current \(I\) into the magnetic field equation gives:
\[ B = \frac{\mu_0 \cdot \frac{q \cdot \omega}{2\pi}}{2r} \]
This simplifies to:
\[ B = \frac{\mu_0 \cdot q \cdot \omega}{4\pi r} \]

### Step 4: Rearranging to find angular velocity \(\omega\)
From the above equation, we can solve for \(\omega\):
\[ \omega = \frac{4\pi r B}{\mu_0 q} \]

### Step 5: Substitute known values
Given:
- \(B = 16 \, \text{weber/m}^2\)
- \(r = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\)
- \(q = 1.6 \times 10^{-19} \, \text{C}\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)

Substituting these values into the equation:
\[ \omega = \frac{4\pi (1 \times 10^{-10}) (16)}{(4\pi \times 10^{-7})(1.6 \times 10^{-19})} \]

### Step 6: Simplifying the expression
The \(4\pi\) terms cancel out:
\[ \omega = \frac{(1 \times 10^{-10})(16)}{(10^{-7})(1.6 \times 10^{-19})} \]
\[ = \frac{16 \times 10^{-10}}{1.6 \times 10^{-26}} = 10^{17} \, \text{rad/s} \]

### Final Answer
Thus, the angular velocity \(\omega\) of the electron is:
\[ \omega = 10^{17} \, \text{rad/s} \]