Step 1: Understanding the Question:
The question asks for the self-inductance ($L$) of an air-core solenoid given its structural dimensions: length, cross-sectional area, and total number of wire turns.
Step 2: Key Formula or Approach:
The self-inductance formula for an ideal long solenoid is given by:
$$L = \frac{\mu_0 N^2 A}{l}$$
where $\mu_0$ is the permeability of free space, $N$ is the total number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.
Step 3: Detailed Explanation:
Let's list the given parameters and convert them into standard SI units:
Length, $l = 31.4 \text{ cm} = 0.314 \text{ m} = 0.1\pi \text{ m}$ (since $\pi \approx 3.14$)
Total turns, $N = 500$
Cross-sectional Area, $A = 10^{-3} \text{ m}^2$
Permeability constant, $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$
Substitute these values directly into our formula:
$$L = \frac{(4\pi \times 10^{-7}) \times (500)^2 \times 10^{-3}}{0.314}$$
Recognizing that $0.314 \approx 0.1\pi$, let's substitute this to simplify the math without tedious decimal divisions:
$$L \approx \frac{4\pi \times 10^{-7} \times 250000 \times 10^{-3}}{0.1\pi}$$
The $\pi$ terms cancel out perfectly:
$$L = \frac{4 \times 10^{-7} \times 2.5 \times 10^5 \times 10^{-3}}{0.1}$$
$$L = \frac{10 \times 10^{-5}}{0.1} = \frac{10^{-4}}{10^{-1}} = 10^{-3} \text{ H}$$
Step 4: Final Answer:
The self-inductance of the solenoid is $10^{-3} \text{ H}$, which matches option (D).