A solenoid of 1000 turns is wound uniformly on a glass tube `50cm` long and `10cm` diameter. The strength of magentic field at the centre of solenoid when a current of `0.1A`. Flows through it is

Options

A. `100 A//m`
B. `200A//m`
C. `400A//m`
D. `50 A//m`

Explanation

To find the strength of the magnetic field at the center of the solenoid, we can use the formula for the magnetic field inside a solenoid:

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

where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) 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:**
- Number of turns \( N = 1000 \)
- Length of the solenoid \( L = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Current \( I = 0.1 \, \text{A} \)

2. **Calculate the number of turns per unit length \( n \):**
\[ n = \frac{N}{L} = \frac{1000}{0.5} = 2000 \, \text{turns/m} \]

3. **Substitute the values into the magnetic field formula:**
\[ B = \mu_0 \cdot n \cdot I \]
Substituting \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \), \( n = 2000 \, \text{turns/m} \), and \( I = 0.1 \, \text{A} \):
\[ B = (4\pi \times 10^{-7}) \cdot (2000) \cdot (0.1) \]

4. **Calculate \( B \):**
\[ B = (4\pi \times 10^{-7}) \cdot (200) = 800\pi \times 10^{-7} \, \text{T} \]
\[ B \approx 800 \times 3.14 \times 10^{-7} \, \text{T} \approx 2.51 \times 10^{-4} \, \text{T} = 0.000251 \, \text{T} \]

5. **Convert to a more convenient unit (if necessary):**
\[ B \approx 0.251 \, \text{mT} \quad (\text{millitesla}) \]

### Final Result:
The strength of the magnetic field at the center of the solenoid when a current of \( 0.1 \, \text{A} \) flows through it is approximately \( 0.251 \, \text{mT} \).