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 of a solenoid**:
The self-inductance \( L \) of a solenoid is given by the formula:
\[ L = \frac{\mu_0 N^2 A}{l} \]
where:
- \( L \) = self-inductance
- \( \mu_0 \) = permeability of free space
- \( N \) = number of turns
- \( A \) = cross-sectional area of the solenoid
- \( l \) = length of the solenoid

2. **Express the number of turns \( N \)**:
The number of turns \( N \) can be expressed in terms of the length of the wire \( x \) and the circumference of the solenoid. The circumference \( C \) of the solenoid is given by:
\[ C = 2\pi r \]
where \( r \) is the radius of the solenoid. Thus, the number of turns \( N \) can be expressed as:
\[ N = \frac{x}{2\pi r} \]

3. **Express the cross-sectional area \( A \)**:
The cross-sectional area \( A \) of the solenoid can be expressed as:
\[ A = \pi r^2 \]

4. **Substitute \( N \) and \( A \) into the self-inductance formula**:
Substitute the expressions for \( N \) and \( A \) into the self-inductance formula:
\[ L = \frac{\mu_0 \left(\frac{x}{2\pi r}\right)^2 \pi r^2}{l} \]

5. **Simplify the equation**:
Simplifying the equation gives:
\[ L = \frac{\mu_0 x^2}{4\pi^2 r^2} \cdot \frac{\pi r^2}{l} = \frac{\mu_0 x^2}{4\pi l} \]

6. **Rearranging for \( x \)**:
Now, rearranging the equation to solve for \( x \):
\[ x^2 = \frac{4\pi l L}{\mu_0} \]
Taking the square root gives:
\[ x = \sqrt{\frac{4\pi l L}{\mu_0}} \]

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