Answer:
[tex]L = 1.88\times10^-^4 H[/tex]
Explanation:
It is given that,
Length of the coil, l = 3 cm = 0.03 m
Area of cross section of the coil, [tex]A=0.5\ cm^2=0.00005\ m^2[/tex]
Number of turns in the coil, N = 300
inductance of the coil is
[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]
[tex]L=\dfrac{4\pi\times 10^{-7}\times (300)^2\times 5\times 10^{-5}}{0.03}\\L = 1.88\times10^-^4 H[/tex]
So, the inductance of the coil is 1.88× 10⁻⁴H
