A toroidal solenoid with an air core has an average radius of 15 cm , area of cross-section `12 "cm"^(2)` and 1200 turns . Ignoring the field variation across the cross-section of the toroid the self-inductance of the toroid is

Options

A. `4.6` mH
B. `6.9` mH
C. `2.3` mH
D. `9.2` mH

Explanation

To find the self-inductance of a toroidal solenoid with the given parameters, we can follow these steps:

### Step 1: Identify the given parameters
- Average radius of the toroid, \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)
- Area of cross-section, \( A = 12 \, \text{cm}^2 = 12 \times 10^{-4} \, \text{m}^2 \)
- Number of turns, \( N = 1200 \)

### Step 2: Calculate the number of turns per unit length
The length of the toroid can be calculated using the formula for the circumference of a circle:
\[ \text{Length} = 2 \pi r = 2 \pi (0.15) \, \text{m} \approx 0.942 \, \text{m} \]
Now, the number of turns per unit length \( n \) is given by:
\[ n = \frac{N}{\text{Length}} = \frac{1200}{0.942} \approx 1273.5 \, \text{turns/m} \]

### Step 3: Use the formula for self-inductance of a toroidal solenoid
The self-inductance \( L \) of a toroidal solenoid is given by the formula:
\[ L = \frac{\mu_0 N^2 A}{2 \pi r} \]
Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ L = \frac{(4\pi \times 10^{-7}) (1200^2) (12 \times 10^{-4})}{2 \pi (0.15)} \]

### Step 5: Simplify the expression
Calculating the components:
1. \( N^2 = 1200^2 = 1440000 \)
2. \( A = 12 \times 10^{-4} \, \text{m}^2 \)
3. \( 2 \pi r = 2 \pi (0.15) \approx 0.942 \)

Now substituting these values:
\[ L = \frac{(4\pi \times 10^{-7}) (1440000) (12 \times 10^{-4})}{0.942} \]

### Step 6: Calculate \( L \)
Calculating the numerator:
\[ 4\pi \times 10^{-7} \times 1440000 \times 12 \times 10^{-4} \approx 2.69 \times 10^{-6} \]
Now divide by \( 0.942 \):
\[ L \approx \frac{2.69 \times 10^{-6}}{0.942} \approx 2.86 \times 10^{-6} \, \text{H} = 2.86 \, \text{mH} \]

### Final Answer
Thus, the self-inductance of the toroidal solenoid is approximately:
\[ L \approx 2.86 \, \text{mH} \]