A toroid of 1000 turns has an average radius of 10 cm and area of cross section `15 cm^(2)`. Calculate the self-inductance of the toroid.

Answer

To calculate the self-inductance of the toroid, we can use the formula for the self-inductance \( L \) of a toroid:

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

Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)
- \( N \) is the number of turns
- \( A \) is the cross-sectional area in square meters
- \( r \) is the average radius in meters

### Step 1: Convert given values to SI units
- Number of turns \( N = 1000 \)
- Average radius \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Cross-sectional area \( A = 15 \, \text{cm}^2 = 15 \times 10^{-4} \, \text{m}^2 = 0.0015 \, \text{m}^2 \)

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

\[ L = \frac{(4\pi \times 10^{-7}) \times (1000)^2 \times (0.0015)}{2 \pi \times 0.1} \]

### Step 3: Simplify the expression
First, calculate \( N^2 \):

\[ N^2 = 1000^2 = 1000000 \]

Now substitute this into the equation:

\[ L = \frac{(4\pi \times 10^{-7}) \times (1000000) \times (0.0015)}{2 \pi \times 0.1} \]

### Step 4: Cancel out \( \pi \)
The \( \pi \) in the numerator and denominator cancels out:

\[ L = \frac{(4 \times 10^{-7}) \times (1000000) \times (0.0015)}{2 \times 0.1} \]

### Step 5: Calculate the numerator and denominator
Calculate the numerator:

\[ 4 \times 10^{-7} \times 1000000 \times 0.0015 = 4 \times 1.5 = 6 \times 10^{-4} \]

Calculate the denominator:

\[ 2 \times 0.1 = 0.2 \]

### Step 6: Final calculation
Now we can calculate \( L \):

\[ L = \frac{6 \times 10^{-4}}{0.2} = 3 \times 10^{-3} \, \text{H} \]

### Step 7: Convert to milliHenries
Since \( 1 \, \text{H} = 1000 \, \text{mH} \):

\[ L = 3 \, \text{mH} \]

### Final Answer
The self-inductance of the toroid is:

\[ L = 3 \, \text{mH} \]

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