The length of a solenoid is `0.1m` and its diameter is very small . A wire is wound over in two layers. The number of turns in the inner layer is 50 and that on the outer layer is 40. The strength of current flowing in two layers in the same direction is 3 ampere. The magnetic induction in the middle of the solenoid will be

Options

A. `3.4xx10^(-3)` Tesla
B. `3.4xx10^(-3)` Gauss
C. `3.4xx10^(3)` Tesla
D. `3.4xx10^(3)` Gauss

Answer

To find the magnetic induction in the middle of the solenoid, we can use the formula for the magnetic field inside a solenoid, which is given by:

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

where:
- \( B \) is the magnetic induction (magnetic field strength),
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( N \) is the total number of turns,
- \( I \) is the current flowing through the solenoid,
- \( L \) is the length of the solenoid.

### Step-by-step Solution:

1. **Calculate the Total Number of Turns (N)**:
- The inner layer has 50 turns and the outer layer has 40 turns.
- Therefore, the total number of turns \( N \) is:
\[ N = 50 + 40 = 90 \]

2. **Identify the Current (I)**:
- The current flowing through both layers is given as \( I = 3 \, \text{A} \).

3. **Identify the Length of the Solenoid (L)**:
- The length of the solenoid is given as \( L = 0.1 \, \text{m} \).

4. **Substitute Values into the Formula**:
- Now we can substitute the values into the formula for magnetic induction:
\[ B = \mu_0 \cdot \frac{N \cdot I}{L} \]
\[ B = (4\pi \times 10^{-7}) \cdot \frac{90 \cdot 3}{0.1} \]

5. **Calculate the Value**:
- First, calculate \( \frac{90 \cdot 3}{0.1} \):
\[ \frac{90 \cdot 3}{0.1} = \frac{270}{0.1} = 2700 \]
- Now substitute this back into the equation:
\[ B = (4\pi \times 10^{-7}) \cdot 2700 \]

6. **Calculate \( 4\pi \cdot 2700 \)**:
- Using \( \pi \approx 3.14 \):
\[ 4\pi \cdot 2700 \approx 4 \cdot 3.14 \cdot 2700 \approx 33912 \]

7. **Combine with the \( 10^{-7} \)**:
- Now, multiply by \( 10^{-7} \):
\[ B \approx 33912 \times 10^{-7} = 3.3912 \times 10^{-3} \, \text{T} \]

8. **Final Result**:
- Therefore, the magnetic induction in the middle of the solenoid is approximately:
\[ B \approx 3.39 \times 10^{-3} \, \text{T} \text{ or } 3.39 \, \text{mT} \]