Question

The self inductance of a solenoid of length $31.4 \text{ cm}$, area of cross section $10^{-3} \text{ m}^2$ having total number of turns $500$ will be nearly [$\mu_0 = 4\pi \times 10^{-7}$ SI unit]

• A. $3 \times 10^{-6} \text{ H}$
• B. $2 \times 10^{-4} \text{ H}$
• C. $0.5 \times 10^{-3} \text{ H}$
• D. $10^{-3} \text{ H}$

Hint:

Whenever a problem in physics uses values like $31.4$, $62.8$, or $0.314$, always rewrite them as multiples of $\pi$ (e.g., $31.4 = 10\pi$). This allows you to instantly cancel out the $\pi$ embedded inside constants like $\mu_0 = 4\pi \times 10^{-7}$, saving precious time.

Answer 1

The Correct Option is D

Step 1: Identify what is asked.
We need the self-inductance $L$ of an air-core solenoid given its length, cross-sectional area and number of turns.
Step 2: Write the solenoid inductance formula.
For a long solenoid, $L = \dfrac{\mu_0 N^2 A}{l}$, where $N$ is the total turns, $A$ the area, and $l$ the length.
Step 3: List the data in SI units.
$l = 31.4\ \text{cm} = 0.314\ \text{m}$, $A = 10^{-3}\ \text{m}^2$, $N = 500$, and $\mu_0 = 4\pi\times 10^{-7}$. A handy trick: $0.314 \approx 0.1\pi$ since $\pi \approx 3.14$.
Step 4: Substitute the values.
\[ L = \frac{(4\pi\times 10^{-7})(500)^2(10^{-3})}{0.1\pi} \]
Step 5: Cancel the $\pi$ terms.
Writing $l = 0.1\pi$ lets $\pi$ cancel top and bottom: \[ L = \frac{4\times 10^{-7}\times 2.5\times 10^{5}\times 10^{-3}}{0.1} \] Here $(500)^2 = 2.5\times 10^5$.
Step 6: Crunch the powers of ten.
The numerator is $4\times 2.5 \times 10^{-7+5-3} = 10\times 10^{-5} = 10^{-4}$. Dividing by $0.1 = 10^{-1}$ gives $L = 10^{-4}/10^{-1} = 10^{-3}\ \text{H}$.
\[ \boxed{L = 10^{-3}\ \text{H}\ \text{(option 4)}} \]