Chapter 32, Problem 14P

To determine

The graph for the self induced emf over the given time interval.

Answer to Problem 14P

The graph for the self induced emf over the given time interval is shown below.

Explanation of Solution

Write the expression to calculate the self induced emf.

    ${\epsilon }_{\text{L}}=-L\frac{dI}{dt}$                                                                                                                (I)

Here, ${\epsilon }_{\text{L}}$ is the self induced emf, $L$ is the inductance and $\frac{dI}{dt}$ is the change in current.

Write the expression to for change in current.

    $\frac{dI}{dt}=\frac{\left(I_2-I_1\right)}{\left(t_2-t_1\right)}$                                                                                                          (II)

Here, $I_2$ is the current at time $t_2$ and $I_1$ is the current at time $t_1$.

Substitute $\frac{\left(I_2-I_1\right)}{\left(t_2-t_1\right)}$ for $\frac{dI}{dt}$ in equation (I).

    ${\epsilon }_{\text{L}}=-L\left(\frac{\left(I_2-I_1\right)}{\left(t_2-t_1\right)}\right)$                                                                                               (III)

Conclusion:

For the interval $0$ to $2\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $-1\text{mA}$ for $I_2$, $0\text{mA}$ for $I_1$, $2\text{ms}$ for $t_2$ and $0\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(-1\text{mA}-0\text{mA}\right)}{\left(2\text{ms}-0\text{ms}\right)}\right)\\ =2.00\text{mV}\end{array}$

For the interval $2\text{ms}$ to $4\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $-2\text{mA}$ for $I_2$, $-1\text{mA}$ for $I_1$, $4\text{ms}$ for $t_2$ and $2\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(-2\text{mA}-\left(-1\text{mA}\right)\right)}{\left(4\text{ms}-2\text{ms}\right)}\right)\\ =2.00\text{mV}\end{array}$

For the interval $4\text{ms}$ to $5.7\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $0\text{mA}$ for $I_2$, $-2\text{mA}$ for $I_1$, $5.7\text{ms}$ for $t_2$ and $4\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(0\text{mA}-\left(-2\text{mA}\right)\right)}{\left(5.7\text{ms}-4\text{ms}\right)}\right)\\ =-4.71\text{mV}\end{array}$

For the interval $5.7\text{ms}$ to $8\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $3\text{mA}$ for $I_2$, $0\text{mA}$ for $I_1$, $8\text{ms}$ for $t_2$ and $5.7\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(3\text{mA}-\left(0\text{mA}\right)\right)}{\left(8\text{ms}-5.7\text{ms}\right)}\right)\\ =-5.22\text{mV}\end{array}$

For the interval $8\text{ms}$ to $10\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $3\text{mA}$ for $I_2$, $3\text{mA}$ for $I_1$, $10\text{ms}$ for $t_2$ and $8\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(3\text{mA}-\left(3\text{mA}\right)\right)}{\left(8\text{ms}-5.7\text{ms}\right)}\right)\\ =0\text{mV}\end{array}$

For the interval $8\text{ms}$ to $10\text{ms}$.

Substitute $4.00\text{mH}$ for $L$, $0\text{mA}$ for $I_2$, $3\text{mA}$ for $I_1$, $12\text{ms}$ for $t_2$ and $1\text{ms}$ for $t_1$ in equation (III) to solve for ${\epsilon }_{\text{L}}$.

    $\begin{array}{c}{\epsilon }_{\text{L}}=-\left(4.00\text{mH}\right)\left(\frac{\left(0\text{mA}-\left(3\text{mA}\right)\right)}{\left(12\text{ms}-10\text{ms}\right)}\right)\\ =6\text{mV}\end{array}$

Therefore, the graph for the self inductance at different time interval is shown in the figure below.

Figure-(1)