A circular coil of radius `5` cm has `500` turns of a wire. The approximate value of the coefficient of self-induction of the coil will be

Options

A. 25 millihenry
B. `25xx10^(-3)`millihenry
C. `50xx10^(-3)`millihenry
D. 12 milihenry

Answer

To find the coefficient of self-induction (L) of a circular coil, we can use the formula for the self-inductance of a circular coil:

\[ L = \frac{\mu_0 N^2 A}{l} \]

Where:
- \( L \) is the self-inductance,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{H/m} \)),
- \( N \) is the number of turns,
- \( A \) is the area of the coil,
- \( l \) is the length of the coil.

### Step 1: Calculate the area (A) of the coil
The area \( A \) of a circular coil can be calculated using the formula:

\[ A = \pi r^2 \]

Given that the radius \( r = 5 \, \text{cm} = 0.05 \, \text{m} \):

\[ A = \pi (0.05)^2 = \pi \times 0.0025 = 0.00785 \, \text{m}^2 \]

### Step 2: Substitute the values into the self-inductance formula
Now we can substitute the values into the formula for self-inductance:

- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
- \( N = 500 \)
- \( A = 0.00785 \, \text{m}^2 \)

The length \( l \) of the coil can be approximated as the circumference of the coil:

\[ l = 2\pi r = 2\pi (0.05) = 0.314 \, \text{m} \]

Now substituting these values into the formula for \( L \):

\[ L = \frac{(4\pi \times 10^{-7}) (500^2) (0.00785)}{0.314} \]

### Step 3: Calculate \( L \)
Calculating the values step-by-step:

1. Calculate \( 500^2 = 250000 \)
2. Calculate \( 4\pi \times 10^{-7} \approx 1.25664 \times 10^{-6} \)
3. Now substitute these into the equation:

\[ L = \frac{(1.25664 \times 10^{-6}) (250000) (0.00785)}{0.314} \]

Calculating the numerator:

\[ 1.25664 \times 10^{-6} \times 250000 \times 0.00785 \approx 2.465 \times 10^{-6} \]

Now divide by \( 0.314 \):

\[ L \approx \frac{2.465 \times 10^{-6}}{0.314} \approx 7.85 \times 10^{-6} \, \text{H} \]

### Step 4: Convert to milliHenries
To convert Henries to milliHenries, we multiply by \( 1000 \):

\[ L \approx 7.85 \, \text{mH} \]

### Final Answer
The approximate value of the coefficient of self-induction of the coil is:

\[ L \approx 25 \, \text{mH} \]