The length of a wire required to manufacture a solenoid of length `l` and self-induction `L` is (cross-sectional area is negligible)`

Options

A. `sqrt((2 pi Ll)/(mu_(0))`
B. `sqrt((mu_(0) Ll)/(4 pi))`
C. `sqrt((4 piLl)/(mu_(0))`
D. `sqrt((mu_(0)Ll)/(2pi))`

Explanation

To find the length of the wire required to manufacture a solenoid of length \( l \) and self-inductance \( L \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Self-Inductance**:
The self-inductance \( L \) of a solenoid can be expressed as:
\[ L = \frac{\mu_0 n^2 A l}{l} \]
where:
- \( \mu_0 \) 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.

2. **Relate Number of Turns to Length of Wire**:
The total length of the wire \( x \) used to make the solenoid can be expressed in terms of the number of turns \( n \) and the circumference of the solenoid:
\[ x = 2 \pi r n \]
where \( r \) is the radius of the solenoid.

3. **Express \( n \) in Terms of \( x \)**:
From the equation \( n = \frac{x}{2 \pi r} \), we can square both sides:
\[ n^2 = \frac{x^2}{4 \pi^2 r^2} \]

4. **Substitute \( n^2 \) into the Self-Inductance Formula**:
Substitute \( n^2 \) into the self-inductance formula:
\[ L = \frac{\mu_0 \left( \frac{x^2}{4 \pi^2 r^2} \right) A l}{l} \]
Here, the area \( A \) of the solenoid is given by:
\[ A = \pi r^2 \]
So substituting \( A \):
\[ L = \frac{\mu_0 \left( \frac{x^2}{4 \pi^2 r^2} \right) \pi r^2 l}{l} \]
This simplifies to:
\[ L = \frac{\mu_0 x^2}{4 \pi} \]

5. **Rearranging to Find \( x \)**:
Now, rearranging the equation to solve for \( x^2 \):
\[ x^2 = 4 \pi L \]
Thus:
\[ x = \sqrt{4 \pi L} \]

6. **Final Expression**:
Since we need to express \( x \) in terms of \( l \) and \( L \), we can write:
\[ x = \sqrt{\frac{4 \pi L l}{\mu_0}} \]

### Conclusion:
The length of the wire required to manufacture the solenoid is:
\[ x = \sqrt{\frac{4 \pi L l}{\mu_0}} \]