A closely wound coil of `100` turns and area of cross-section `1cm^(3)` has a coefficient of self-induction `1 mH`. The magnetic induction in the centre of the core of the coil when a current of `2A` flows in it, will be

Options

A. `0.022 Wb m^(-2)`
B. `0.4 Wb m^(-2)`
C. `0.8 Wb M^(-2)`
D. `1 Wb m^(-2)`

Explanation

To find the magnetic induction in the center of a closely wound coil when a current flows through it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns (N) = 100 turns
- Area of cross-section (A) = 1 cm² = \(1 \times 10^{-4} \, \text{m}^2\)
- Coefficient of self-induction (L) = 1 mH = \(1 \times 10^{-3} \, \text{H}\)
- Current (I) = 2 A

2. **Calculate the Radius of the Coil:**
- The area of a circle is given by the formula:
\[ A = \pi r^2 \]
- Rearranging for radius (r):
\[ r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{1 \times 10^{-4}}{\pi}} \approx \sqrt{3.183 \times 10^{-5}} \approx 5.64 \times 10^{-3} \, \text{m} \]

3. **Use the Formula for Magnetic Field (B) at the Center of the Coil:**
- The formula for the magnetic field at the center of a coil is:
\[ B = \frac{\mu_0 N I}{2r} \]
- Where \(\mu_0\) (permeability of free space) is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

4. **Substitute the Values into the Formula:**
- Plugging in the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 100 \times 2}{2 \times (5.64 \times 10^{-3})} \]

5. **Simplify the Expression:**
- The \(2\) in the numerator and denominator cancels out:
\[ B = \frac{(4\pi \times 10^{-7}) \times 100}{(5.64 \times 10^{-3})} \]
- Calculate:
\[ B \approx \frac{(4 \times 3.14 \times 10^{-7} \times 100)}{(5.64 \times 10^{-3})} \]
\[ B \approx \frac{1.256 \times 10^{-5}}{5.64 \times 10^{-3}} \approx 0.00223 \, \text{T} \]

6. **Final Result:**
- The magnetic induction (B) in the center of the coil when a current of 2 A flows through it is approximately:
\[ B \approx 0.00223 \, \text{T} \text{ or } 2.23 \, \text{mT} \]