A solenoid of 500 turns has a length of \( 50 \, \text{cm} \) and a cross-sectional area of \( 4 \times 10^{-4} \, \text{m}^2 \). If the permeability of free space is \( 4\pi \times 10^{-7} \, \text{T·m/A} \), what is the self-inductance of the solenoid?
The self-inductance (\( L \)) of a solenoid is calculated using the formula:
\[ L = \mu_0 \cdot \frac{N^2 \cdot A}{l} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} \): Permeability of free space
- \( N = 500 \): Number of turns
- \( A = 4 \times 10^{-4} \, \text{m}^2 \): Cross-sectional area
- \( l = 50 \, \text{cm} = 0.5 \, \text{m} \): Length of the solenoid
Substitute the given values:
\[ L = \left( 4\pi \times 10^{-7} \right) \cdot \frac{\left( 500 \right)^2 \cdot \left( 4 \times 10^{-4} \right)}{0.5} \]
First, calculate \( 500^2 \):
\[ 500^2 = 250,000 \]
Substitute into the formula:
\[ L = \left( 4\pi \times 10^{-7} \right) \cdot \frac{250,000 \cdot 4 \times 10^{-4}}{0.5} \]
Simplify \( 250,000 \cdot 4 \times 10^{-4} \):
\[ 250,000 \cdot 4 \times 10^{-4} = 100 \]
Now substitute into the formula:
\[ L = \left( 4\pi \times 10^{-7} \right) \cdot \frac{100}{0.5} \]
Simplify \( \frac{100}{0.5} \):
\[ \frac{100}{0.5} = 200 \]
Now calculate \( L \):
\[ L = \left( 4\pi \times 10^{-7} \right) \cdot 200 \]
\[ L = 800\pi \times 10^{-7} \]
Substitute \( \pi \approx 3.14 \):
\[ L = 800 \cdot 3.14 \times 10^{-7} \]
\[ L = 2512 \times 10^{-7} \, \text{H} \]
\[ L = 2.512 \times 10^{-4} \, \text{H} = 0.2512 \, \text{mH} \]
The self-inductance of the solenoid is:
\[ L = 0.2512 \, \text{mH} \]
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