A thin solenoid of length `0.4 m` and having `500` turns of wire carries a current `1A`,then find the magnetic field on the axis inside the solenoid.

Explanation

To find the magnetic field inside a thin solenoid, we can use the formula:

\[ B = \mu_0 \cdot n \cdot I \]

where:
- \( B \) is the magnetic field inside the solenoid,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( n \) is the number of turns per unit length of the solenoid,
- \( I \) is the current flowing through the solenoid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the solenoid, \( L = 0.4 \, \text{m} \)
- Number of turns, \( N = 500 \)
- Current, \( I = 1 \, \text{A} \)

2. **Calculate the number of turns per unit length (n):**
\[ n = \frac{N}{L} = \frac{500}{0.4} = 1250 \, \text{turns/m} \]

3. **Substitute the values into the magnetic field formula:**
\[ B = \mu_0 \cdot n \cdot I \]
Substituting \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \), \( n = 1250 \, \text{turns/m} \), and \( I = 1 \, \text{A} \):
\[ B = (4\pi \times 10^{-7}) \cdot (1250) \cdot (1) \]

4. **Calculate the magnetic field:**
\[ B = 4\pi \times 1250 \times 10^{-7} \]
\[ B = 5000\pi \times 10^{-7} \, \text{T} \]
\[ B \approx 5 \times 10^{-4} \, \text{T} \quad (\text{using } \pi \approx 3.14) \]

5. **Final Result:**
The magnetic field inside the solenoid is approximately:
\[ B \approx 5 \times 10^{-4} \, \text{T} \]