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 based on the self-inductance of the first coil and the number of turns in each coil. Here’s the step-by-step solution:

### Step 1: Understand the relationship between self-inductance and number of turns
The self-inductance \( L \) of a coil is directly proportional to the square of the number of turns \( N \) in the coil. This relationship can be expressed mathematically as:
\[ L \propto N^2 \]
This means that if we have two coils, the ratio of their self-inductances can be given by:
\[ \frac{L_1}{L_2} = \left(\frac{N_1}{N_2}\right)^2 \]
where \( L_1 \) and \( L_2 \) are the self-inductances of the first and second coils, and \( N_1 \) and \( N_2 \) are the number of turns in the first and second coils, respectively.

### Step 2: Identify the known values
From the problem, we have:
- For the first coil:
- Number of turns \( N_1 = 600 \)
- Self-inductance \( L_1 = 108 \, \text{mH} \)
- For the second coil:
- Number of turns \( N_2 = 500 \)
- Self-inductance \( L_2 = ? \)

### Step 3: Set up the equation using the known values
Using the relationship established in Step 1, we can set up the equation:
\[ \frac{L_1}{L_2} = \left(\frac{N_1}{N_2}\right)^2 \]
Substituting the known values:
\[ \frac{108 \, \text{mH}}{L_2} = \left(\frac{600}{500}\right)^2 \]

### Step 4: Simplify the right side of the equation
Calculating the right side:
\[ \frac{600}{500} = 1.2 \]
\[ \left(1.2\right)^2 = 1.44 \]
So the equation becomes:
\[ \frac{108 \, \text{mH}}{L_2} = 1.44 \]

### Step 5: Solve for \( L_2 \)
Now, rearranging the equation to solve for \( L_2 \):
\[ L_2 = \frac{108 \, \text{mH}}{1.44} \]
Calculating \( L_2 \):
\[ L_2 = 75 \, \text{mH} \]

### Final Answer
The self-inductance of the second coil is \( L_2 = 75 \, \text{mH} \).

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