Answer:
2.36 μ H
Explanation:
Given,
Number of turns= 90
diameter = 1.3 cm = 0.013 m
unscratched length = 57 cm = 0.57 m
Area, A = π r²
= π x 0.0065² = 1.32 x 10⁻⁴ m²
we know,
[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]
[tex]L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}[/tex]
L = 2.36 μ H
Hence, the inductance of the unstretched cord is equal to 2.36 μ H
