The self inductance of a solenoid of length L, area of cross-section A and having N turns is-

Options

A. `(mu_(0)N^(2)S)/(L)`
B. `(mu_(0)NS)/(L)`
C. `mu_(0)N^(2)LS`
D. `mu_(0)NLS`

Explanation

To find the self-inductance \( L \) of a solenoid with length \( L \), cross-sectional area \( A \), and \( N \) turns, we can follow these steps:

### Step 1: Understand the Magnetic Field in the Solenoid
The magnetic field \( B \) inside a long solenoid can be expressed as:
\[ B = \mu_0 \cdot n \cdot I \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length of the solenoid, given by \( n = \frac{N}{L} \),
- \( I \) is the current flowing through the solenoid.

### Step 2: Calculate the Total Magnetic Flux
The total magnetic flux \( \Phi \) through the solenoid can be calculated using:
\[ \Phi = N \cdot B \cdot A \]
Substituting \( B \) into the equation, we get:
\[ \Phi = N \cdot (\mu_0 \cdot n \cdot I) \cdot A \]
Replacing \( n \) with \( \frac{N}{L} \):
\[ \Phi = N \cdot \left(\mu_0 \cdot \frac{N}{L} \cdot I\right) \cdot A \]

### Step 3: Substitute the Area
The cross-sectional area \( A \) can be expressed as:
\[ A = \pi r^2 \]
Thus, we can rewrite the flux as:
\[ \Phi = \mu_0 \cdot \frac{N^2}{L} \cdot I \cdot A \]

### Step 4: Relate Flux to Inductance
The self-inductance \( L \) is defined by the relationship:
\[ \Phi = L \cdot I \]
By comparing the two expressions for \( \Phi \), we can equate them:
\[ L \cdot I = \mu_0 \cdot \frac{N^2}{L} \cdot I \cdot A \]

### Step 5: Solve for Self-Inductance
Dividing both sides by \( I \) (assuming \( I \neq 0 \)):
\[ L = \mu_0 \cdot \frac{N^2 \cdot A}{L} \]
Thus, the self-inductance \( L \) of the solenoid is given by:
\[ L = \frac{\mu_0 \cdot N^2 \cdot A}{L} \]

### Final Answer
The self-inductance of a solenoid of length \( L \), area of cross-section \( A \), and having \( N \) turns is:
\[ L = \frac{\mu_0 \cdot N^2 \cdot A}{L} \]

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