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A solenoid has 4000 turns over it with inductance of 0.126 H. The energy stored in the magnetic field, when a current of 2 A flows in the solenoid will be
1. 0.252 J
2. 8000 J
3. 0.504 J
4. 1008 J

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Correct Answer - Option 1 : 0.252 J

Concept:

The inductance of a solenoid is given by:

\(L = \frac{{{\mu _0}{N^2}A}}{l} \)

N = Number of turns

A = Area of the solenoid

l = length of the solenoid

Also, the energy stored by the inductor is given by:

\(E=\frac{1}{2}LI^2\)

Calculation:

Given that, inductance (L) = 0.126 H

Current (I) = 2 A

\(E = \frac{1}{2}L{I^2} = \frac{1}{2} \times 0.126 \times {2^2} \)

= 0.252 J

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