A solenoid has fixed N number of turns and fixed radius `a` its length is given by `l` which can be varied its self-inductance is proportional to

Options

A. `l`
B. `(1)/(l)`
C. `l^(2)`
D. `(1)/(l^(2))`

Answer

To determine how the self-inductance \( L \) of a solenoid varies with its length \( l \), we can follow these steps:

### Step 1: Understand the formula for self-inductance
The self-inductance \( L \) of a solenoid can be expressed as:
\[ L = \frac{n \Phi}{I} \]
where \( n \) is the number of turns per unit length, \( \Phi \) is the magnetic flux, and \( I \) is the current.

### Step 2: Determine the magnetic field inside the solenoid
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n I \]
where \( \mu_0 \) is the permeability of free space and \( n \) can be expressed as:
\[ n = \frac{N}{l} \]
with \( N \) being the total number of turns and \( l \) being the length of the solenoid.

### Step 3: Substitute \( n \) into the magnetic field equation
Substituting \( n \) into the equation for \( B \):
\[ B = \mu_0 \frac{N}{l} I \]

### Step 4: Calculate the magnetic flux \( \Phi \)
The magnetic flux \( \Phi \) through one turn of the solenoid is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the cross-sectional area of the solenoid. For a solenoid with radius \( a \):
\[ A = \pi a^2 \]
Thus,
\[ \Phi = B \cdot \pi a^2 = \left(\mu_0 \frac{N}{l} I\right) \cdot \pi a^2 \]

### Step 5: Substitute \( \Phi \) into the self-inductance formula
Now substituting \( \Phi \) back into the self-inductance formula:
\[ L = \frac{N \left(\mu_0 \frac{N}{l} I \cdot \pi a^2\right)}{I} \]
The current \( I \) cancels out:
\[ L = \frac{\mu_0 N^2 \pi a^2}{l} \]

### Step 6: Analyze the relationship between \( L \) and \( l \)
From the final expression for \( L \):
\[ L \propto \frac{1}{l} \]
This indicates that the self-inductance \( L \) is inversely proportional to the length \( l \) of the solenoid.

### Conclusion
Thus, the self-inductance of the solenoid is proportional to \( \frac{1}{l} \).