Answer:
Inductance as calculated is 13.12 mH
Solution:
As per the question:
Length of the coil, l = 12 cm = 0.12 m
Diameter, d = 1.7 cm = 0.017 m
No. of turns, N = 235
Now,
Area of cross-section of the wire, A = [tex]\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}[/tex]
We know that the inductance of the coil is given by the formula:
[tex]L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH[/tex]
