Question

A coil of wire of radius 'r' has 600 turns and a self-inductance of 108 mH. The self-inductance of a coil with same radius and 500 turns is ______.

• A. 80 mH
• B. 75 mH
• C. 108 mH
• D. 90 mH

Hint:

Self-inductance is essentially a measure of magnetic "inertia". Because every turn creates magnetic flux that cuts through every other turn, the effect compounds quadratically ($N \times N$), not linearly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are comparing the self-inductance of two coils. The geometric shape (radius) is identical, but the number of turns has been changed from 600 to 500. We must find the new inductance.

Step 2: Key Formula or Approach:
The formula for the self-inductance $L$ of a standard circular coil or solenoid is:
$$L = \frac{\mu_0 N^2 A}{l}$$
From this, it is evident that self-inductance is directly proportional to the square of the number of turns, provided all other geometric factors (radius, area, length) remain constant.
$$L \propto N^2 \implies \frac{L_2}{L_1} = \left(\frac{N_2}{N_1}\right)^2$$

Step 3: Detailed Explanation:
Given values:
$L_1 = 108 \text{ mH}$, $N_1 = 600$
$L_2 = ?$, $N_2 = 500$
Set up the proportion:
$$\frac{L_2}{108} = \left( \frac{500}{600} \right)^2$$
Simplify the fraction inside the parenthesis:
$$\frac{L_2}{108} = \left( \frac{5}{6} \right)^2$$
$$\frac{L_2}{108} = \frac{25}{36}$$
Isolate $L_2$:
$$L_2 = 108 \times \frac{25}{36}$$
Observe that $108 / 36 = 3$:
$$L_2 = 3 \times 25 = 75 \text{ mH}$$

Step 4: Final Answer:
The new self-inductance is 75 mH, matching option (b).