A coil of `Cu` wire (radius `r`, self-inductance `L`) is bent in two concentric turns each having radius `(r )/(2)`. The self-inductance now

Options

A. `2L`
B. `L`
C. `4L`
D. `L//2`

Answer

To solve the problem, we need to determine the self-inductance of a coil after it has been bent into two concentric turns. Here’s a step-by-step solution:

### Step 1: Understand the relationship between self-inductance, number of turns, and radius
The self-inductance \( L \) of a coil is given by the formula:
\[ L \propto n^2 \cdot r \]
where \( n \) is the number of turns and \( r \) is the radius of the coil.

### Step 2: Identify the initial conditions
Initially, we have:
- Self-inductance before bending: \( L_1 = L \)
- Radius before bending: \( r_1 = r \)
- Number of turns before bending: \( n_1 = 1 \) (since it's a single coil)

### Step 3: Identify the new conditions after bending
After bending the coil into two concentric turns:
- New radius: \( r_2 = \frac{r}{2} \)
- New number of turns: \( n_2 = 2 \) (since it is now two turns)

### Step 4: Write the relationship for self-inductance after bending
Using the relationship for self-inductance, we can write:
\[ \frac{L_1}{L_2} = \frac{n_1^2 \cdot r_1}{n_2^2 \cdot r_2} \]

### Step 5: Substitute the known values
Substituting the values we have:
\[ \frac{L}{L_2} = \frac{(1)^2 \cdot r}{(2)^2 \cdot \left(\frac{r}{2}\right)} \]
This simplifies to:
\[ \frac{L}{L_2} = \frac{r}{4 \cdot \frac{r}{2}} = \frac{r}{2r} = \frac{1}{2} \]

### Step 6: Solve for \( L_2 \)
From the above equation, we can express \( L_2 \):
\[ L_2 = 2L \]

### Conclusion
The self-inductance after bending the coil is:
\[ L_2 = 2L \]

### Final Answer
The correct option is option number 1: \( 2L \).

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