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. `2pi xx 10^(12)` rad/sec
D. `4pi xx 10^(12)` rad/sec

Answer

To find the angular velocity of an electron revolving around a proton that produces a magnetic field of 16 Weber/m² in a circular orbit of radius 1 Å (1 x 10⁻¹⁰ m), we can follow these steps:

### Step 1: Understand the relationship between magnetic field, current, and angular velocity
The magnetic field \( B \) produced by a current \( I \) in a circular loop is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( R \) is the radius of the orbit.

### Step 2: Relate current to charge and time period
The current \( I \) can be expressed in terms of charge \( Q \) and time period \( T \):
\[ I = \frac{Q}{T} \]
For one complete revolution, the time period \( T \) is related to the angular velocity \( \omega \) as:
\[ T = \frac{2\pi}{\omega} \]

### Step 3: Substitute \( I \) in the magnetic field equation
Substituting \( I \) into the magnetic field equation gives:
\[ B = \frac{\mu_0 \cdot \frac{Q}{T}}{2R} = \frac{\mu_0 Q \cdot \omega}{2 \cdot 2\pi R} \]
Rearranging this, we find:
\[ B = \frac{\mu_0 Q \cdot \omega}{4\pi R} \]

### Step 4: Solve for angular velocity \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{B \cdot 4\pi R}{\mu_0 Q} \]

### Step 5: Substitute known values
We know:
- \( B = 16 \, \text{Wb/m}^2 \)
- \( R = 1 \times 10^{-10} \, \text{m} \)
- \( Q = 1.6 \times 10^{-19} \, \text{C} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

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

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

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

### Conclusion
The correct option is **Option 1: \( 10^{17} \, \text{rad/s} \)**.