A solenoid of 2.5 m length and 2.0 cm diameter possesses 10 turns per cm. A current of 0.5 A is flowing through it . The magnetic induction at axis inside the solenoid is

Options

A. `2pixx10^(-4)T`
B. `2pixx10^(-5)T`
C. `2pixx10^(-6)T`
D. `2pixx10^(-7)T`

Answer

To find the magnetic induction (B) at the axis inside a solenoid, we can use the formula:

\[ B = \mu_0 n I \]

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Length of the solenoid, \( L = 2.5 \, \text{m} \)
- Diameter of the solenoid, \( d = 2.0 \, \text{cm} \) (not needed for this calculation)
- Turns per unit length, \( n = 10 \, \text{turns/cm} \)
- Current, \( I = 0.5 \, \text{A} \)

2. **Convert turns per centimeter to turns per meter**:
\[ n = 10 \, \text{turns/cm} = 10 \times 100 \, \text{turns/m} = 1000 \, \text{turns/m} \]

3. **Substitute the values into the formula**:
\[ B = \mu_0 n I \]
\[ B = (4\pi \times 10^{-7} \, \text{T m/A}) \times (1000 \, \text{turns/m}) \times (0.5 \, \text{A}) \]

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

5. **Final Result**:
\[ B \approx 6.2832 \times 10^{-4} \, \text{T} \quad (\text{since } 2\pi \approx 6.2832) \]

Thus, the magnetic induction at the axis inside the solenoid is approximately \( 2\pi \times 10^{-4} \, \text{T} \).