Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of magnetic field at the centre of the coil?

Answer

To find the magnitude of the magnetic field at the center of a tightly wound coil, we can use the formula for the magnetic field at the center of a circular coil. The formula for the magnetic field \( B \) at the center of a circular coil with \( N \) turns, carrying a current \( I \), and having a radius \( R \) is given by:

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

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \( N = 100 \)
- Current, \( I = 1 \, \text{A} \)
- Radius, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Substitute the values into the formula:**
\[ B = \frac{\mu_0 N I}{2R} = \frac{(4\pi \times 10^{-7}) \times 100 \times 1}{2 \times 0.1} \]

3. **Calculate the denominator:**
- \( 2R = 2 \times 0.1 = 0.2 \)

4. **Now substitute this back into the equation:**
\[ B = \frac{(4\pi \times 10^{-7}) \times 100}{0.2} \]

5. **Calculate \( 100/0.2 \):**
- \( 100/0.2 = 500 \)

6. **Now substitute this back into the equation:**
\[ B = 4\pi \times 10^{-7} \times 500 \]

7. **Calculate \( 4\pi \times 500 \):**
- \( 4\pi \times 500 = 2000\pi \)

8. **Now, using \( \pi \approx 3.14 \):**
\[ B \approx 2000 \times 3.14 \times 10^{-7} \]
\[ B \approx 6280 \times 10^{-7} \, \text{T} \]

9. **Convert to scientific notation:**
\[ B \approx 6.28 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field at the center of the coil is approximately \( 6.28 \times 10^{-4} \, \text{T} \).