A current of 4.0 A is maintained in a coil of self-inductance 8.0 mH. The energy stored in the coil is:

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  1. 64 mJ
  2. 128 mJ
  3. 32 mJ
  4. 16 mJ

Answer (Detailed Solution Below)

Option 1 : 64 mJ
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Concept:

Self-inductance:

  • Self-inductance is the property of the current-carrying coil that resists or opposes the change of current flowing through it.

  • This occurs mainly due to the self-induced emf produced in the coil itself.

  • The emf induced in the coil is given by, \(e=-L\frac{dI}{dt}\) where L = self-inductance

  • The energy stored in the coil is, \(E=\frac{1}{2}LI^2\) where I = current

Calculation:

Given,

The current, I = 4.0 A, self-inductance of a coil, L = 8.0 mH

The energy stored in the coil is calculated by the given formula,

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

\(E=\frac{1}{2}\times 8\times 10^{-3}\times 4^2 \)

E = 64 mJ

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