The expression for magnetic induction inside a solenoid of length `L` carrying a current `I` and having `N` number of turns is

Options

A. `mu_0n/4πl`
B. `mu_0l/4πn`
C. `mu_0/4πnl`
D. `mu_0nl`

Answer

To derive the expression for magnetic induction (magnetic field) inside a solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Parameters**:
- Let \( L \) be the length of the solenoid.
- Let \( I \) be the current flowing through the solenoid.
- Let \( N \) be the total number of turns of the wire in the solenoid.

2. **Calculate the Number of Turns per Unit Length**:
- The number of turns per unit length \( n \) can be calculated as:
\[ n = \frac{N}{L} \]
This gives us the density of turns in the solenoid.

3. **Use the Formula for Magnetic Field Inside a Solenoid**:
- The magnetic field \( B \) inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
where \( \mu_0 \) is the permeability of free space.

4. **Substitute the Value of \( n \)**:
- Substitute the expression for \( n \) from step 2 into the formula from step 3:
\[ B = \mu_0 \left(\frac{N}{L}\right) I \]

5. **Final Expression**:
- Rearranging gives us the final expression for the magnetic induction inside the solenoid:
\[ B = \frac{\mu_0 N I}{L} \]

### Conclusion:
The expression for magnetic induction inside a solenoid of length \( L \) carrying a current \( I \) and having \( N \) number of turns is:
\[ B = \frac{\mu_0 N I}{L} \]