Question

A coil with an inductance of 2.0H and a resistance of$\mathbf{10}\mathit{\Omega }$is suddenly connected to an ideal battery with$\mathit{\epsilon }\mathbf{=}\mathbf{100}\mathit{V}$. (a) What is the equilibrium current? (b) How much energy is stored in the magnetic field when this current exists in the coil?

Short answer

1. The equilibrium current is $i_0=10A$
2. The amount of energy is stored in the magnetic field when the current exists in the coil $U_B=100\mathrm{J}$

Step-by-step solution

Given

1. The inductance of a coil is L = 2.0 H
2. The resistance of a coil is $R=10\Omega$
3. The emf of an ideal battery is $\epsilon =100V$

Understanding the concept

We need to use the equation of Ohm’s law to calculate the equilibrium current and the equation of energy stored in the magnetic field of an inductor.

Formulae:

$$\begin{gathered} i_0=\frac{E}{R} \\ {U}_B=\frac{1}{2}L{i}_0^2 \end{gathered}$$

(a) Calculate the equilibrium current

According to Ohm’s law, the equilibrium current through the coil is

$$\begin{gathered} i_0=\frac{\epsilon }{R} \\ {i}_0=\frac{100V}{10\Omega } \\ {i}_0=10A \end{gathered}$$

(b) Calculate the amount of energy is stored in the magnetic field when the current exists in the coil

The expression for the magnetic energy is

$$\begin{gathered} U_B=\frac{1}{2}L{i}_0^2 \\ {U}_B=\frac{1}{2}\times 2.0\mathrm{H}\times {\left(10\mathrm{A}\right)}^2{U}_B=100\mathrm{J} \end{gathered}$$

Amount of energy stored in the magnetic field is 100J