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[tex]length \: (\( l \)) = 0.70 mm = 0.70 \times 10^{-3} m,[/tex]
[tex]- Cross-sectional area (\( A \)) = 1500 mm\(^2\) = 1500 \times 10^{-6} m\(^2\),[/tex]
[tex]- Permeability (\( \mu \)) = 4500 µWb/At·m = 4500 \times 10^{-6} Wb/At·m.[/tex]
[tex]\[ \mathcal{R} = \frac{0.70 \times 10^{-3} \text{ m}}{4500 \times 10^{-6} \text{ Wb/At·m} \times 1500 \times 10^{-6} \text{ m}^2} \]
[/tex]
[tex]\[ \mathcal{R} = \frac{0.70 \times 10^{-3}}{4500 \times 1500 \times 10^{-12}} \]
[/tex]
[tex]\[ \mathcal{R} = \frac{0.70 \times 10^{-3}}{6.75 \times 10^{-6}} \][/tex]
[tex]\[ \mathcal{R} = \frac{0.70}{6.75} \times 10^{3} \][/tex]
[tex]\[ \mathcal{R} \approx 0.1037 \times 10^{3} \][/tex]
[tex]\[ \mathcal{R} {\boxed{\approx 103.7 \text{ At/Wb} \]}}[/tex]