A circular coil of wire consisting of `100` turns each of radius `8cm` carries a current of `0.4A.` What is the magnitude of magnetic field at the center of the coil

Answer

To find the magnitude of the magnetic field at the center of a circular coil of wire, we can use the formula for the magnetic field due to a coil with multiple turns. The formula is:

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

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns, \( n = 100 \)
- Radius of the coil, \( R = 8 \, \text{cm} = 0.08 \, \text{m} \) (convert cm to m)
- Current, \( I = 0.4 \, \text{A} \)

2. **Substitute the Values into the Formula:**
- We know \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
- Substitute \( n \), \( \mu_0 \), \( I \), and \( R \) into the formula:
\[ B = \frac{100 \times (4\pi \times 10^{-7}) \times 0.4}{2 \times 0.08} \]

3. **Calculate the Denominator:**
- Calculate \( 2R \):
\[ 2R = 2 \times 0.08 = 0.16 \, \text{m} \]

4. **Calculate the Numerator:**
- Calculate \( n \mu_0 I \):
\[ n \mu_0 I = 100 \times (4\pi \times 10^{-7}) \times 0.4 = 160\pi \times 10^{-7} \]

5. **Combine the Results:**
- Now substitute back into the equation for \( B \):
\[ B = \frac{160\pi \times 10^{-7}}{0.16} \]

6. **Simplify:**
- Divide \( 160 \) by \( 0.16 \):
\[ \frac{160}{0.16} = 1000 \]
- Thus,
\[ B = 1000\pi \times 10^{-7} = 10^{-4}\pi \, \text{T} \]

7. **Final Calculation:**
- Using \( \pi \approx 3.14 \):
\[ B \approx 3.14 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field at the center of the coil is approximately \( 3.14 \times 10^{-4} \, \text{T} \).