The self inductance L of a solenoid depends on the number of turns per unit length 'n' as

Options

A. `L oo n`
B. `L oo n^(2)`
C. `L oo n^(-1)`
D. `L oo n^(-2)`

Answer

To determine how the self-inductance \( L \) of a solenoid depends on the number of turns per unit length \( n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnetic Flux**:
The magnetic flux \( \Phi \) through the solenoid can be expressed as:
\[ \Phi = n \cdot B \cdot A \]
where \( B \) is the magnetic field inside the solenoid and \( A \) is the cross-sectional area.

2. **Magnetic Field in a Solenoid**:
The magnetic field \( B \) inside a long solenoid is given by:
\[ B = \mu_0 \cdot n \cdot I \]
where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current flowing through the solenoid.

3. **Cross-Sectional Area**:
The cross-sectional area \( A \) of the solenoid can be expressed as:
\[ A = \pi r^2 \]
where \( r \) is the radius of the solenoid.

4. **Substituting for Magnetic Flux**:
Now, substituting \( B \) and \( A \) into the equation for magnetic flux:
\[ \Phi = n \cdot (\mu_0 \cdot n \cdot I) \cdot (\pi r^2) \]
This simplifies to:
\[ \Phi = \mu_0 \cdot n^2 \cdot \pi r^2 \cdot I \]

5. **Relating Magnetic Flux to Self-Inductance**:
The self-inductance \( L \) is defined by the relationship:
\[ \Phi = L \cdot I \]
By comparing this with our expression for \( \Phi \), we can equate:
\[ L \cdot I = \mu_0 \cdot n^2 \cdot \pi r^2 \cdot I \]

6. **Solving for Self-Inductance**:
Dividing both sides by \( I \) (assuming \( I \neq 0 \)):
\[ L = \mu_0 \cdot n^2 \cdot \pi r^2 \]
This shows that the self-inductance \( L \) is directly proportional to the square of the number of turns per unit length \( n \).

### Conclusion:
Thus, we conclude that the self-inductance \( L \) of a solenoid depends on the number of turns per unit length \( n \) as:
\[ L \propto n^2 \]