Due to `10` ampere of current flowing in a circular coil of `10 cm` radius, the magnetic field produced at its centre is `3.14xx10^(-3) Weber//m^(3)`. The number of turns in the coil will be

Options

A. `5000`
B. `100`
C. `50`
D. `25`

Explanation

To find the number of turns in a circular coil given the current, radius, and magnetic field at its center, we can use the formula derived from the Biot-Savart law. The magnetic field \( B \) at the center of a circular coil with \( n \) turns is given by:

\[ B = \frac{\mu_0 n I}{2R} \]

Where:
- \( B \) = magnetic field at the center of the coil
- \( \mu_0 \) = permeability of free space \( = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( n \) = number of turns
- \( I \) = current in amperes
- \( R \) = radius of the coil in meters

### Step 1: Convert the radius to meters
Given the radius \( R = 10 \, \text{cm} = 0.1 \, \text{m} \).

### Step 2: Substitute known values into the formula
We know:
- \( B = 3.14 \times 10^{-3} \, \text{T} \)
- \( I = 10 \, \text{A} \)
- \( R = 0.1 \, \text{m} \)

Substituting these values into the formula:

\[ 3.14 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) n (10)}{2(0.1)} \]

### Step 3: Simplify the equation
First, simplify the right side:

\[ 3.14 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) n (10)}{0.2} \]

This can be rewritten as:

\[ 3.14 \times 10^{-3} = (20\pi \times 10^{-7}) n \]

### Step 4: Solve for \( n \)
Now, isolate \( n \):

\[ n = \frac{3.14 \times 10^{-3}}{20\pi \times 10^{-7}} \]

### Step 5: Calculate \( n \)
Calculating the denominator:

\[ 20\pi \approx 62.83 \quad \text{(using } \pi \approx 3.14\text{)} \]

Now substituting this value back into the equation:

\[ n = \frac{3.14 \times 10^{-3}}{62.83 \times 10^{-7}} = \frac{3.14}{62.83} \times 10^{4} \]

Calculating \( \frac{3.14}{62.83} \):

\[ n \approx 0.05 \times 10^{4} = 5 \times 10^{2} = 50 \]

### Final Answer
The number of turns in the coil is \( n = 50 \).