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Answer:
where:
- \( L \) is the inductance,
- \( \mu \) is the permeability of the core,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area of the core,
- \( l \) is the length of the core.
Given:
- Inductance (\( L \)) = 30 mH = 30 \times 10^{-3} H,
- Permeability (\( \mu \)) = 1.26 \times 10^{-6} H/m,
- Cross-sectional area (\( A \)) = 100 mm\(^2\) = 100 \times 10^{-6} m\(^2\),
- Length (\( l \)) = 0.05 m.
Rearranging the formula to solve for \( N \):
\[ N^2 = \frac{L \cdot l}{\mu \cdot A} \]
[tex] \large {Plugging \: in \: the \: given values}[/tex]
[tex]\[ N^2 = \frac{30 \times 10^{-3} \text{ H} \times 0.05 \text{ m}}{1.26 \times 10^{-6} \text{ H/m} \times 100 \times 10^{-6} \text{ m}^2} \][/tex]
[tex]\[ N^2 = \frac{1.5 \times 10^{-3}}{1.26 \times 10^{-6} \times 100 \times 10^{-6}} \]
[/tex]
[tex]\[ N^2 = \frac{1.5 \times 10^{-3}}{1.26 \times 10^{-12}} \][/tex]
[tex]\[ N^2 = \frac{1.5}{1.26} \times 10^9 \][/tex]
[tex]\[ N^2 \approx 1.1905 \times 10^9 \][/tex]
[tex]\[ N \approx \sqrt{1.1905 \times 10^9} \]
[/tex]
[tex] \: \red{ \boxed{\[ N \approx 34,510 \]}}[/tex]