A coil of wire of a certain radius has `600` turns and a self-inductance of `108 mH`. The self-inductance of a `2^(nd)` similar coil of `500` turns will be

Options

A. `74 mh`
B. `75 mH`
C. `76 mH`
D. `77 mH`

Explanation

To solve the problem, we need to find the self-inductance of a second coil with a different number of turns, given the self-inductance of the first coil. Here's the step-by-step solution:

### Step 1: Understand the relationship of self-inductance with the number of turns
The self-inductance \( L \) of a coil is proportional to the square of the number of turns \( N \):
\[ L \propto N^2 \]
This means that if we have two coils with self-inductances \( L_1 \) and \( L_2 \) and number of turns \( N_1 \) and \( N_2 \), we can express the relationship as:
\[ \frac{L_1}{L_2} = \frac{N_1^2}{N_2^2} \]

### Step 2: Assign the known values
From the problem, we have:
- For the first coil:
- \( N_1 = 600 \) turns
- \( L_1 = 108 \, \text{mH} = 108 \times 10^{-3} \, \text{H} \)

- For the second coil:
- \( N_2 = 500 \) turns
- \( L_2 = ? \)

### Step 3: Set up the equation using the relationship
Using the relationship established in Step 1, we can write:
\[ \frac{L_1}{L_2} = \frac{N_1^2}{N_2^2} \]
Substituting the known values:
\[ \frac{108 \times 10^{-3}}{L_2} = \frac{600^2}{500^2} \]

### Step 4: Calculate \( \frac{600^2}{500^2} \)
Calculating the squares:
\[ 600^2 = 360000 \quad \text{and} \quad 500^2 = 250000 \]
Thus,
\[ \frac{600^2}{500^2} = \frac{360000}{250000} = \frac{36}{25} = 1.44 \]

### Step 5: Substitute back to find \( L_2 \)
Now substituting back into the equation:
\[ \frac{108 \times 10^{-3}}{L_2} = 1.44 \]
Rearranging gives:
\[ L_2 = \frac{108 \times 10^{-3}}{1.44} \]

### Step 6: Calculate \( L_2 \)
Calculating \( L_2 \):
\[ L_2 = \frac{108 \times 10^{-3}}{1.44} \approx 75 \times 10^{-3} \, \text{H} = 75 \, \text{mH} \]

### Final Answer
The self-inductance of the second coil with 500 turns is approximately \( 75 \, \text{mH} \).

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