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 find the magnetic field inside a 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 (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns per unit length of the solenoid,
- \( I \) is the current flowing through the solenoid.

### Step 1: Identify the given values
- Number of turns, \( N = 100 \)
- Length of the solenoid, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 0.5 \, \text{A} \)

### Step 2: Calculate the number of turns per unit length \( n \)
\[ n = \frac{N}{L} \]
Substituting the values:
\[ n = \frac{100}{0.1} = 1000 \, \text{turns/m} \]

### Step 3: Substitute the values into the magnetic field formula
Now we can substitute \( n \) and \( I \) into the formula for \( B \):
\[ B = \mu_0 \cdot n \cdot I \]
Substituting \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \):
\[ B = (4\pi \times 10^{-7}) \cdot (1000) \cdot (0.5) \]

### Step 4: Calculate \( B \)
\[ B = 4\pi \times 10^{-7} \times 1000 \times 0.5 \]
\[ B = 2\pi \times 10^{-4} \, \text{T} \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ B \approx 2 \times 3.14 \times 10^{-4} \approx 6.28 \times 10^{-4} \, \text{T} \]

### Final Answer
The magnetic field inside the solenoid is approximately:
\[ B \approx 6.28 \times 10^{-4} \, \text{T} \]