$A$ current vs. time graph of the current passing through a solenoid is shown in the figure. For which time is the back electromotive force $(emf)$ a maximum? If the back $emf$ at $t = 3\;s$ is $e$, find the back $emf$ at $t = 7\;s$, $15\;s$, and $40\;s$. $OA$, $AB$, and $BC$ are straight line segments.

(D) The back $emf$ in a solenoid is given by $\varepsilon = -L \frac{dI}{dt}$, where $L$ is the self-inductance and $\frac{dI}{dt}$ is the slope of the current-time graph.
$1$. At $t = 3\;s$ (segment $OA$): The slope is $\frac{1 - 0}{5 - 0} = 0.2\;A/s$. Thus, $e = -L(0.2) = -0.2L$. Therefore, $L = -5e$.
$2$. At $t = 7\;s$ (segment $AB$): The slope is $\frac{-2 - 1}{10 - 5} = \frac{-3}{5} = -0.6\;A/s$. The back $emf$ is $\varepsilon_1 = -L(-0.6) = 0.6L$. Substituting $L = -5e$, we get $\varepsilon_1 = 0.6(-5e) = -3e$.
$3$. At $t = 15\;s$ (segment $BC$): The slope is $\frac{0 - (-2)}{30 - 10} = \frac{2}{20} = 0.1\;A/s$. The back $emf$ is $\varepsilon_2 = -L(0.1) = -0.1L$. Substituting $L = -5e$, we get $\varepsilon_2 = -0.1(-5e) = 0.5e = \frac{e}{2}$.
$4$. At $t = 40\;s$: The current is constant at $0\;A$ (beyond $t=30\;s$), so the slope $\frac{dI}{dt} = 0$. Thus, the back $emf$ is $0$.
The magnitude of the back $emf$ is maximum when the slope $|\frac{dI}{dt}|$ is maximum. Comparing slopes: $|0.2|$, $|-0.6|$, and $|0.1|$. The maximum slope is $|-0.6|$ occurring between $t = 5\;s$ and $t = 10\;s$.