A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.

Answer

To calculate the magnetic field at the center of a solenoid along its axis, we can use the formula:

\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]

where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(N\) is the total number of turns,
- \(L\) is the 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, \(L = \pi \, \text{m}\)
- Diameter of the solenoid = 5 cm (not needed for this calculation)
- Number of turns, \(N = 1000\)
- Current, \(I = 5 \, \text{A}\)

2. **Calculate the value of \(N/L\):**
\[ \frac{N}{L} = \frac{1000}{\pi} \]

3. **Substitute the values into the formula:**
\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]
\[ B = (4\pi \times 10^{-7}) \cdot \left(\frac{1000}{\pi}\right) \cdot 5 \]

4. **Simplify the expression:**
- The \(\pi\) in the numerator and denominator cancels out:
\[ B = 4 \times 10^{-7} \cdot 1000 \cdot 5 \]
\[ B = 20 \times 10^{-4} \, \text{T} \]

5. **Convert to standard form:**
\[ B = 2 \times 10^{-3} \, \text{T} \]

### Final Answer:
The magnetic field at the center along the axis of the solenoid is \(2 \times 10^{-3} \, \text{T}\) (Tesla).