In Figure (a), the inductor has 25 turns and the ideal battery has an emf of 16 V. Figure (b) gives the magnetic flux$\mathbf{\phi }$ through each turn versus the current i through the inductor. The vertical axis scale is set by ${\mathbf{\phi }}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{Tm}}^{\mathbf{2}}$, and the horizontal axis scale is set by ${\mathbf{i}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{A}$If switch S is closed at time t = 0, at what rate$\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}$will the current be changing at$\mathit{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}{\mathit{\tau }}_{\mathbf{L}}$?

i) Number of turns is$N=25$ .

ii) Battery emf is$\epsilon =16V$ .

iii) Figure 30-63 is the RL circuit.

iv) Figure 30-64 is the graph of flux Vs current.

v) Y scale of graph is${\varphi }_s=4.0\times {10}^{-4}T.m^2$

vi) X scale of graph is$i_s=2.00A$

We use the concept of inductance and current through RL circuit. First we find inductance from the number of turns and the graph. Using the equation, we can find its derivative, which will be the rate of change of current. For the given value of time, we can find the rate of change of current.

Formulae:

$$\begin{gathered} L=\frac{N\varphi }{i} \\ i=\frac{V}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \end{gathered}$$

We can find the inductance from the given value and graph:

$L=\frac{N\varphi }{i}$

From thegraph, we can find$\frac{\varphi }{i}$ :

$$\begin{gathered} \frac{{\Phi }_s}{i_s}=\frac{4.0\times {10}^{-4}}{2.00}=2.0\times {10}^{-4} \\ L=25\times 2.00\times {10}^{-4}=5.0\times {10}^{-3} \end{gathered}$$

We find the derivative of the equation:

$$\begin{gathered} i=\frac{V}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ \frac{di}{dt}=\frac{V}{R}\left(-\frac{1}{{\tau }_L}\right)\left(e^{-\frac{t}{{\tau }_L}}\right) \end{gathered}$$

We know ; we can write

$$\begin{gathered} \frac{di}{dt}=\frac{V}{R}\left(\frac{R}{L}\right)\left(e^{-\frac{t}{{\tau }_L}}\right) \\ \frac{di}{dt}=\frac{V}{L}\left(e^{-\frac{t}{{\tau }_L}}\right) \end{gathered}$$

Plugging the values, we get

$$\begin{gathered} \frac{di}{dt}=\frac{16}{5.0\times {10}^{-3}}\left(e^{-\frac{t}{{\tau }_L}}\right) \\ \frac{di}{dt}=3200\left(e^{-1.5}\right) \\ \frac{di}{dt}=7.1\times {10}^{2}A/s \end{gathered}$$