SSC JE Electrical Engineering 24 Jan 2018 Morning Paper

Question : 129 of 199

Marks: +1, -0

Determine the value of self–inductance (in mH) of a 6–centimetre long air–cored solenoid, if the coil has 240 turns and the diameter of the coil is 4 cm.

1.212
1.514
1.614
1.625

Validate

Solution:

l = 6 cm =

6 × 10^{–2}

m
N = 240
d = 4 cm ⇒ r = 2 cm =

2 × 10^{–2}

m.
L = ?
Now self inductance (L)

={\µ_0N^2A}/\l

={4π×10^{-7}×(240)^2×A}/{6×10^{-2}}

Now

A = πr^2

= π × (2 × 10^{–2})^2

= π × 4 × 10^{–4}

∴L={4π×10^{-7}×240^2×π×4×10^{-4}}/{6×10^{-2}}

= 1.514 mH

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