Question

A coil of inductance L is carrying steady current I. What is the nature of its steady energy?
A. Magnetic
B. Electrical
C. Both magnetic and electrical
D. Heat

Answer 1

Hint: This problem is based on the basics of electrical circuits. What is necessary to know is the potential energy values for different electrical components such as potential energy value for inductor is $\dfrac{1}{2}L{{I}^{2}}$, for a capacitor is $\dfrac{{{Q}^{2}}}{2C}$ and for a resistor is, ${{I}^{2}}R$ and how do these energies dissipate as.

Complete step by step solution:
Let’s start by understanding how the power dissipation process occurs in a resistor of resistance (R). When a charge (q) moves through a resistor of resistance (R), it loses qV amount of potential energy, where the drop in potential is (V) across the resistor. This potential energy gets converted to Heat known as heat dissipation process. Hence, for the steady current I flowing through the resistor (R), the energy dissipated is $VI$ or ${{I}^{2}}R$. This process of heat dissipation by a resistor is used in electrical appliances such as regular iron, immersion rods etc.
Now, let’s try to understand the energy stored in a capacitor of capacitance (C). If this capacitor was initially uncharged and gained an amount of charge (q), this causes the potential difference (V) between the plates. Hence, q=CV. The work done is given by,\[dW=Vdq=\dfrac{q}{C}dq\]. Hence the total work done is,$W=\int_{o}^{q}{\dfrac{q}{C}dq}=\dfrac{{{q}^{2}}}{2C}.$ This is the work done in delivering a charge q, to the capacitor, which is equal to the energy stored by the capacitor. Hence, energy stored by the capacitor is $\dfrac{{{q}^{2}}}{2C}=\dfrac{1}{2}C{{V}^{2}}$. This is known as Electrical potential energy, since a charged capacitor, in the absence of a potential source, can act as a potential source. It stores the electrical potential energy. The capacitors are usually used in multiple electronic components such as fans, tube lights etc.
Moving on to the Inductors, let’s consider an Inductor of inductance (L) and (I) amount of steady current flowing through the inductor. The power of the circuit is given by$P=LI\dfrac{dI}{dt}$. Hence, the energy stored in an inductor is given by,\[E=\int{Pdt=}\int{LI\dfrac{dI}{dt}dt=\int{LIdI=\dfrac{1}{2}L{{I}^{2}}.}}\] Thus, the magnetic potential energy stored by an inductor is $\dfrac{1}{2}L{{I}^{2}}$. The various places where inductors can be seen are in transformer components, chokes and relay circuits etc. Hence Option A is correct.

Note: An easy way of remembering that the steady state energy of an inductor is of magnetic form is by remembering it as the way an inductor is, the winding form of an inductor causes magnetic flux to be produced. Hence, the electrical energy gets converted into magnetic energy by an inductor.
You might remember how a nail coiled by a wire around it gets converted into a magnet, when the wire is connected to a battery. This coiled wire is an example of how an inductor works and how electrical energy gets converted to magnetic energy.