Chapter 31, Problem 1P

To determine

The average induced emf in the circuit for the given time interval.

Answer to Problem 1P

The average induced emf for the given time interval is $\text{100}\text{V}$.

Explanation of Solution

Given Info: The inductance of the given inductor is $2.00\text{H}$, steady current is $0.500\text{A}$ and time interval is $10.0\text{ms}$.

Formula to calculate induced emf is,

$\epsilon =L\frac{di}{dt}$ (1)

Here,

$\epsilon$ is the induced emf in the circuit.

$L$ is the inductance of inductor.

$\frac{di}{dt}$ is the rate of change in the current.

Negative sign indicates the induced emf is opposite to the direction of current.

The expression for the rate of change of current is,

$\frac{di}{dt}=\frac{i_2-i_1}{t}$ (2)

Substitute $0.500\text{A}$ for $i_2$, $0\text{A}$ for $i_1$, $10\text{ms}$ for $t$. in equation (2).

$\frac{di}{dt}=\frac{0.500\text{A}-0\text{A}}{10\text{ms}}$

$\begin{array}{c}\frac{di}{dt}=\frac{0.500\text{A}}{10\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)}\\ =\frac{0.500\text{A}}{10\times {10}^{-3}\text{s}}\\ =50\text{A}/\text{s}\end{array}$

Substitute $2.00\text{H}$ for $L$ and $50\text{A}/\text{s}$ for $\frac{di}{dt}$ in equation (1) to calculate induced emf.

$\begin{array}{c}\epsilon =L\frac{di}{dt}\\ =2.00\text{H}\times 50\text{A}/\text{s}\\ =100\text{V}\end{array}$

Conclusion:

Therefore, the induced emf in the inductor during the given time interval is $100\text{V}$.