SSC JE Electrical Engineering 22 Jan 2018 Morning Paper

Question : 129 of 200

Marks: +1, -0

Determine the self-inductance (in mH) of a 3 m long air-cored solenoid, when the coil has 300 turns and the diameter of the coil 12 cm.

Solution:

Given data
N = 300 turns
d = 12 cm ⇒ r = 6cm = 0.06 m
l = 3 m
Now self inductance is given by

M={\µ_0\µ_r\N^2\A}/{\l}

={4π×10^{-17}×300^2×3.14×0.06^2}/3

∴M=0.42 \mH

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