A solenoid consists of 100 turns of wire and has a length of 10 cm. The magnetic field inside the solenoid when it carries a current of 0.5 A will be :

Options

A. `6.28xx10^(-4)T`
B. `6.28xx10^(-5)T`
C. `3.14xx10^(-4)T`
D. None of these

Explanation

To solve the problem of finding the magnetic field inside a solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns (N) = 100
- Length of the solenoid (L) = 10 cm = 0.1 m (convert to meters)
- Current (I) = 0.5 A

2. **Calculate the Number of Turns per Unit Length (n):**
- The number of turns per unit length (n) is given by the formula:
\[ n = \frac{N}{L} \]
- Substituting the values:
\[ n = \frac{100}{0.1} = 1000 \, \text{turns/m} \]

3. **Use the Formula for Magnetic Field (B) inside the Solenoid:**
- The formula for the magnetic field inside a solenoid is:
\[ B = \mu_0 n I \]
- Where \(\mu_0\) (the permeability of free space) is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

4. **Substitute the Values into the Formula:**
- Now substituting the values of \(\mu_0\), \(n\), and \(I\):
\[ B = (4\pi \times 10^{-7}) \times (1000) \times (0.5) \]

5. **Calculate the Magnetic Field:**
- First, calculate \(1000 \times 0.5 = 500\).
- Now substitute this back into the equation:
\[ B = 4\pi \times 10^{-7} \times 500 \]
- This simplifies to:
\[ B = 2000\pi \times 10^{-7} \]

6. **Approximate the Value of \(\pi\):**
- Using \(\pi \approx 3.14\):
\[ B \approx 2000 \times 3.14 \times 10^{-7} = 6280 \times 10^{-7} = 6.28 \times 10^{-4} \, \text{T} \]

7. **Final Answer:**
- The magnetic field inside the solenoid when it carries a current of 0.5 A is:
\[ B \approx 6.28 \times 10^{-4} \, \text{T} \]