A coil having 800 turns, area of cross section `12 cm^(2)` and 15 cm long is connected in series with another coil having 500 turns, area of cross section `10 cm^(2)` and 10 cm long. Calculate the net inductance of the system.

Answer

To solve the problem of finding the net inductance of two coils connected in series, we can follow these steps:

### Step 1: Identify the given values for both coils.
- For Coil 1:
- Number of turns, \( N_1 = 800 \)
- Area of cross-section, \( A_1 = 12 \, \text{cm}^2 = 12 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Length, \( L_1 = 15 \, \text{cm} = 0.15 \, \text{m} \) (conversion from cm to m)

- For Coil 2:
- Number of turns, \( N_2 = 500 \)
- Area of cross-section, \( A_2 = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Length, \( L_2 = 10 \, \text{cm} = 0.10 \, \text{m} \) (conversion from cm to m)

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

### Step 3: Calculate the inductance \( L_1 \) for Coil 1.
Substituting the values for Coil 1 into the formula:
\[ L_1 = \mu_0 \frac{N_1^2 A_1}{L_1} = 4\pi \times 10^{-7} \frac{(800)^2 \times (12 \times 10^{-4})}{0.15} \]

Calculating \( L_1 \):
\[ L_1 = 4\pi \times 10^{-7} \frac{640000 \times 12 \times 10^{-4}}{0.15} \]
\[ = 4\pi \times 10^{-7} \frac{7680 \times 10^{-4}}{0.15} \]
\[ = 4\pi \times 10^{-7} \times 51200 \times 10^{-4} \]
\[ = 4\pi \times 10^{-7} \times 0.512 \approx 6.43 \times 10^{-7} \, \text{H} \]

### Step 4: Calculate the inductance \( L_2 \) for Coil 2.
Substituting the values for Coil 2 into the formula:
\[ L_2 = \mu_0 \frac{N_2^2 A_2}{L_2} = 4\pi \times 10^{-7} \frac{(500)^2 \times (10 \times 10^{-4})}{0.10} \]

Calculating \( L_2 \):
\[ L_2 = 4\pi \times 10^{-7} \frac{250000 \times 10 \times 10^{-4}}{0.10} \]
\[ = 4\pi \times 10^{-7} \frac{2500 \times 10^{-4}}{0.10} \]
\[ = 4\pi \times 10^{-7} \times 25000 \times 10^{-4} \]
\[ = 4\pi \times 10^{-7} \times 0.25 \approx 3.14 \times 10^{-7} \, \text{H} \]

### Step 5: Calculate the net inductance \( L \) of the system.
For two coils in series, the net inductance \( L \) is given by:
\[ L = \sqrt{L_1 \times L_2} \]

Substituting the values:
\[ L = \sqrt{(6.43 \times 10^{-7}) \times (3.14 \times 10^{-7})} \]
\[ = \sqrt{2.018 \times 10^{-13}} \approx 4.49 \times 10^{-7} \, \text{H} \]

### Final Answer:
The net inductance of the system is approximately \( 4.49 \, \text{mH} \).