A point charge is moving in a circular path of radius `10cm` with angular frequency `40pi` ra//s. The magnetic field produced by it at the center is `3.8xx10^(-10)T`. Then the value of charge is:

Options

A. `2xx10^(-6)C`
B. `3xx10^(-5)C`
C. `4xx10^(-5)C`
D. `4xx10^(-6)C`

Answer

To find the value of the charge moving in a circular path, we can use the formula for the magnetic field produced by a point charge moving in a circular motion. The magnetic field \( B \) at the center of the circular path is given by:

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{q \cdot \omega}{r} \]

Where:
- \( B \) is the magnetic field at the center,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( q \) is the charge,
- \( \omega \) is the angular frequency,
- \( r \) is the radius of the circular path.

### Step 1: Write down the known values
Given:
- \( B = 3.8 \times 10^{-10} \, \text{T} \)
- \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( \omega = 40\pi \, \text{rad/s} \)

### Step 2: Substitute the known values into the formula
Substituting the known values into the magnetic field formula:

\[ 3.8 \times 10^{-10} = \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{q \cdot (40\pi)}{0.1} \]

### Step 3: Simplify the equation
The \( 4\pi \) terms cancel out:

\[ 3.8 \times 10^{-10} = 10^{-7} \cdot \frac{q \cdot (40\pi)}{0.1} \]

This simplifies to:

\[ 3.8 \times 10^{-10} = 10^{-7} \cdot (400\pi q) \]

### Step 4: Isolate \( q \)
Now, we can isolate \( q \):

\[ q = \frac{3.8 \times 10^{-10}}{10^{-7} \cdot 400\pi} \]

### Step 5: Calculate \( q \)
Calculating the right-hand side:

\[ q = \frac{3.8 \times 10^{-10}}{4 \times 10^{-5} \cdot \pi} \]

Using \( \pi \approx 3.14 \):

\[ q \approx \frac{3.8 \times 10^{-10}}{1.256 \times 10^{-4}} \approx 3.02 \times 10^{-6} \, \text{C} \]

### Step 6: Round off the answer
Rounding off, we find:

\[ q \approx 3 \times 10^{-6} \, \text{C} \]

### Conclusion
Thus, the value of the charge is:

\[ \boxed{3 \times 10^{-6} \, \text{C}} \]