SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 137 of 200

Marks: +1, -0

The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying current of 10 A is :

Solution:

∵

L={µ_0N^2A}/l={µ_0N^2×πr^2}/l

where

µ_0=4π×10^{-7}

wb/m
∴

L={4π×10^{-7}×1000×1000×π×3/2×3/2×10^{-4}}/{30×10^{-2}}

= 29.43 ×10^{-4}

Henry.
Now stored energy

(E) =1/2LI^2

∴

E=1/2×29.43×10^{-4}×(10)^2

= 0.14715 J
∴ Option (c) is correct.

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