The length of a thin wire required to manufacture a solenoid of length`l = 100 cm` and inductance `L = 1 mH`, if the solenoid's cross-sectional diameter is considerably less than its length is

Options

A. `1 km`
B. `0.10 km`
C. `0.010 km`
D. `10 km`

Explanation

To solve the problem of finding the length of a thin wire required to manufacture a solenoid with given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Values:**
- Length of the solenoid, \( l = 100 \, \text{cm} = 1 \, \text{m} \)
- Inductance, \( L = 1 \, \text{mH} = 1 \times 10^{-3} \, \text{H} \)

2. **Use the Formula for Inductance of a Solenoid:**
The 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 (\( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \))
- \( n \) is the number of turns per unit length
- \( A \) is the cross-sectional area of the solenoid
- \( l \) is the length of the solenoid

3. **Relate Number of Turns to Length of Wire:**
The total length of wire \( L_1 \) used in the solenoid can be expressed as:
\[ L_1 = n \cdot l \]
where \( n \) is the total number of turns.

4. **Calculate the Cross-Sectional Area \( A \):**
For a solenoid with a circular cross-section, the area \( A \) can be given as:
\[ A = \pi r^2 \]
where \( r \) is the radius of the solenoid.

5. **Substituting Values into the Inductance Formula:**
Rearranging the inductance formula gives:
\[ L = \frac{\mu_0 (L_1/l)^2 \cdot A}{l} \]
Substituting \( A = \pi r^2 \) into the equation, we get:
\[ L = \frac{\mu_0 L_1^2 \pi r^2}{l^3} \]

6. **Solving for Length of Wire \( L_1 \):**
Rearranging for \( L_1 \):
\[ L_1 = l \sqrt{\frac{L \cdot l^2}{\mu_0 \pi r^2}} \]

7. **Assuming a Small Diameter:**
Since the diameter is considerably less than the length, we can assume a small radius \( r \) for simplification. However, we will not need the radius for the calculation since we are looking for \( L_1 \) in terms of \( l \) and \( L \).

8. **Substituting Known Values:**
Plugging in the values:
\[ L_1 = 1 \, \text{m} \cdot \sqrt{\frac{1 \times 10^{-3} \cdot (1)^2}{4\pi \times 10^{-7}}} \]

9. **Calculating the Length of Wire:**
\[ L_1 = 1 \cdot \sqrt{\frac{1 \times 10^{-3}}{4\pi \times 10^{-7}}} \]
\[ L_1 = 1 \cdot \sqrt{\frac{10^4}{4\pi}} \approx 1 \cdot \sqrt{10^4} \approx 100 \, \text{m} \]

10. **Convert to Kilometers:**
\[ L_1 = 100 \, \text{m} = 0.1 \, \text{km} \]

### Final Answer:
The length of the thin wire required to manufacture the solenoid is **0.10 km**.