A solenoid of length `l` metre has self-inductance `L` henry. If number of turns are doubled, its self-inductance

Options

A. (a)Remains same
B. (b)Becomes 2L henry
C. (c)Becomes 4L henry
D. (d)Becomes `L/(sqrt(2))` henry

Answer

To solve the problem regarding the self-inductance of a solenoid when the number of turns is doubled, we will follow these steps:

### Step 1: Understand the formula for self-inductance
The self-inductance \( L \) of a solenoid is given by the formula:
\[ L = \frac{\mu_0 n^2 A}{l} \]
where:
- \( L \) is the self-inductance,
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length,
- \( A \) is the cross-sectional area of the solenoid,
- \( l \) is the length of the solenoid.

### Step 2: Identify the initial conditions
Let:
- The initial number of turns be \( n \),
- The initial self-inductance be \( L \).

### Step 3: Determine the new number of turns
If the number of turns is doubled, the new number of turns becomes:
\[ n' = 2n \]

### Step 4: Substitute the new number of turns into the self-inductance formula
Now, substituting \( n' \) into the self-inductance formula:
\[ L' = \frac{\mu_0 (n')^2 A}{l} = \frac{\mu_0 (2n)^2 A}{l} \]

### Step 5: Simplify the expression for the new self-inductance
Calculating \( (2n)^2 \):
\[ (2n)^2 = 4n^2 \]
Thus, the new self-inductance becomes:
\[ L' = \frac{\mu_0 (4n^2) A}{l} = 4 \left(\frac{\mu_0 n^2 A}{l}\right) = 4L \]

### Step 6: Conclusion
Therefore, when the number of turns is doubled, the new self-inductance \( L' \) is:
\[ L' = 4L \]

### Final Answer
The self-inductance becomes four times the initial self-inductance.

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