Question

68. II A inductor with negligible resistance has a 1.0 A current through it. The current starts to increase at $t=0\text{s}$, creating a constant $5.0\text{mV}$voltage across the inductor. How much charge passes through the inductor between $5.0\text{mV}$and $t=5.0\text{s}$?

Short answer

The charge is .$22\text{C}$

Step-by-step solution

Introduction

Faraday's law states that the electromotive force developed in a conducting loop is equal to the rate of change of magnetic flux inside the loop with respect to time. The expression for the potential difference across the inductor is:

$\Delta V_L=-L\frac{dI}{dt}$

Where,$L$is the inductance and$\frac{dI}{dt}$is the rate of change of current

Step 2. Explanation

The inductance is $3.6\text{mH}$and the potential difference is$5.0\text{mV}$.

The expression for voltage across the inductor is:

$\begin{array}{l}\Delta V_L=-L\frac{dI}{dt}\\ \frac{dI}{dt}=\frac{\Delta V_L}{L}\\ =\frac{5.0\text{mV}}{3.6\text{mH}}=1.389\text{A}/\text{s}\end{array}$

The current as a function of time is:

$\begin{array}{l}I\left(t\right)=I_e+\left(\frac{dI}{dt}\right)t\\ =I_e+\frac{\Delta V}{L}t\end{array}$

$Q=\int \left(I\right(t\left)\right)dt$

Thus the charge is$=$22

Step 2. Explanation

The inductance is $3.6\text{\hspace{0.33em}mH}$and the potential difference is $5.0\text{\hspace{0.33em}mV}$

The expression for voltage across the inductor is:

$\begin{array}{l}\Delta V_L=-L\frac{dI}{dt}\\ \frac{dI}{dt}=\frac{\Delta V_L}{L}\\ =\frac{5.0\text{mV}}{3.6\text{mH}}=1.389\text{A}/\text{s}\end{array}$

The current as a function of time is:

$\begin{array}{l}I\left(t\right)=I_e+\left(\frac{dI}{dt}\right)t\\ =I_e+\frac{\Delta V}{L}t\end{array}$

Also, the charge is the integral of the current function.

$Q=\int \left(I\right(t\left)\right)dt$

Thus the charge is:

$\begin{array}{l}Q={\int }_0^t\left(I_e+\frac{\Delta V}{L}t\right)dt\\ ={\left[I_{e}t+\frac{1}{2}\frac{\Delta V}{L}{t}^2\right]}_0^{5.0\text{s}}\\ =I_e\left(5.0\text{s}\right)+\frac{1}{2}\left(1.389\text{A}/\text{s}\right){\left(5.0\text{s}\right)}^2\\ =22\text{\hspace{0.33em}C}\end{array}$

Hence the charge is$22\text{C}$