Question

An inductor having coefficient of self-induction $40\,mH$ . What is the energy stored in it when a current of $2\,A$ is passed through it?
A. $40\,mJ$
B. $80\,mJ$
C. $20\,mJ$
D. $100\,mJ$

Answer 1

We start by defining the terms mentioned in the question. We then write down the given information. Then we proceed to find the formula to find the energy stored in an inductor. We will be able to find out the solution of this question by direct substitution of the given values into the formula.

Formulas used:
Energy stored in an inductor is given by the formula,
$U = \dfrac{1}{2}L{i^2}$
Where, $i$ is the current passing through the inductor and $L$ is the coefficient of self-inductance.

Complete step by step answer:
Self-inductance is defined as the property of an electric circuit that permits self-induction.Let us start by gathering the data given to us;
The value of coefficient of self-inductance is given as, $L = 40\,mH$.
The value of current is given as, $i = 2A$.
Using the formula to find the energy stored in an inductor,
$U = \dfrac{1}{2}L{i^2}$
Substituting the given values in the equation, we get

\Rightarrow U= \dfrac{1}{2} \times 40 \times {10^{ - 3}} \times {2^2} \\\ \Rightarrow U= 80 \times {10^{ - 3}}J \\\ \therefore U= 80\,mJ$$ We convert $$mH$$ into $$H$$ so that the value we end up with is in joules.Therefore, the value of energy stored in this inductor would be $$80\,mJ$$. **Hence, option B is the right answer.** **Note:** Self-inductance is characterized as the induction of a voltage in a current-conveying wire when the current in the actual wire is evolving. On account of self-inductance, the magnetic field made by a changing current in the actual circuit incites a voltage in the same circuit. Self-induction is also known as the inertia of electricity. This is because the coil that has self-induction has the property by virtue of which it maintains the magnetic field in it and opposes the change of flux by inducing current.