Q31P - Introduction to Electrodynamics

Question

Suppose the circuit in Fig. 7.41 has been connected for a long time when suddenly, at time $\mathbf{t}\mathbf{=}\mathbf{0}$, switch S is thrown from A to B, bypassing the battery.

Notice the similarity to Eq. 7.28-in a sense, the rectangular toroid is a short coaxial cable, turned on its side.

 

(a) What is the current at any subsequent time t? 

(b) What is the total energy delivered to the resistor? 

(c) Show that this is equal to the energy originally stored in the inductor.

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The} \mathrm{energy} \mathrm{delivered} \mathrm{to} \mathrm{the} \mathrm{resistor} \mathrm{is}\frac{1}{2}\mathrm{L}{\left(\frac{{\mathrm{\epsilon }}_0}{\mathrm{R}}\right)}^2 . \\ \left(\mathrm{b}\right)\mathrm{The} \mathrm{energy} \mathrm{delivered} \mathrm{to} \mathrm{the} \mathrm{resistor} \mathrm{is} \frac{1}{2}\mathrm{L}{\left(\frac{{\mathrm{\epsilon }}_0}{\mathrm{R}}\right)}^2 . \\ \left(\mathrm{c}\right)\mathrm{The} \mathrm{energy} \mathrm{delivered} \mathrm{to} \mathrm{the} \mathrm{inductor} \mathrm{is} \frac{1}{2}\mathrm{L}{\left(\frac{{\mathrm{\epsilon }}_0}{\mathrm{R}}\right)}^2 . \end{gathered}$$

Step 1: Energy in magnetic fields

When the current is supplied to a circuit in a magnetic field, then it moves against the direction of the back emf. 

 

The work done by a charge in moving against the back emf of the circuit is described as the ‘energy in the magnetic field’.

Step 2(a): The current at any subsequent time

Using Ohm’s law, the formula for the initial current of the circuit is given by,

$I_0=\frac{{\epsilon }_0}{R}$

 

The formula for the Induced emf in the circuit is given by,

 

$$\begin{gathered} -L\frac{dI}{dt}=IR \\ \frac{dI}{dt}=-\frac{R}{L}I \end{gathered}$$

 

Then the expression for the current in the circuit as a function of the subsequent time is given by,

 

$$\begin{gathered} I=I_0{e}^{\frac{-Rt}{L}} \\ I\left(t\right)=\frac{{\epsilon }_0}{R}{e}^{\frac{-Rt}{L}} \end{gathered}$$

 

Hence, the current at any subsequent time is $I\left(t\right)=\frac{{\epsilon }_0}{R}{e}^{\frac{-Rt}{L}}$.

Step 3(b): The total energy delivered to the resistor

The formula for the power due to the resistance of the circuit is given by,

 $$\begin{gathered} P=I^{2}R \\ P={\left(\frac{{\epsilon }_0}{R}e\frac{-Rt}{L}\right)}^{2}R \\ P={\left(\frac{{\epsilon }_0}{R}\right)}^2{e}^{\frac{-2Rt}{L}}R \\ P=\left(\frac{{{\epsilon }_0}^2}{R^2}\right)e^{\frac{-2Rt}{L}}R \end{gathered}$$

Rewrite the equation as:

$P=\left(\frac{{{\epsilon }_0}^2}{R}\right)e^{\frac{-2Rt}{L}}$

Similarly, the formula for the power required by the charge to move against the emf is given by,

 $$\begin{gathered} P=\frac{dW}{dt} \\ dW=Pdt \\ dW=\left(\frac{{{\epsilon }_0}^2}{R}\right)e^{\frac{-2Rt}{L}}dt \end{gathered}$$

Here, W is the work done or the energy delivered to the resistor.

Integrating both sides,

$$\begin{gathered} W=\left(\frac{{{\epsilon }_0}^2}{R}\right){\int }_0^{\infty }{e}^{\frac{-2Rt}{L}}dt \\ W=\left(\frac{{{\epsilon }_0}^2}{R}\right){\left(-\frac{L}{2R}{e}^{\frac{-2Rt}{L}}\right)}_0^{\infty } \\ W=\left(\frac{{{\epsilon }_0}^2}{R}\right)\left(0+\frac{L}{2R}\right) \\ W=\frac{1}{2}L{\left(\frac{{\epsilon }_0}{R}\right)}^2 \end{gathered}$$

Hence, the energy delivered to the resistor is $\frac{1}{2}L{\left(\frac{{\epsilon }_0}{R}\right)}^2$.

Step 4(c): The energy originally stored in the inductor

Using the energy formula, the expression for the energy originally stored in the inductor is given by,

 

$$\begin{gathered} W_0=\frac{1}{2}L{I}_0^2 \\ {W}_0=\frac{1}{2}L{\left(\frac{{\epsilon }_0}{R}\right)}^2 \end{gathered}$$

 

Comparing the energy delivered to the resistor with the energy originally stored in the inductor,

 

$W=W_0$

 

Hence, the energy delivered to the resistor is equal to the energy originally stored in the inductor.