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

$\varepsilon =-\frac{d(N\phi _{B})}{dt}$

$\varepsilon =-L\frac{dl}{dt}$

Thus, negative sign indicates that induced emf (e) opposes any change (increase or decrease) of current in the coil.

When the rate of change of current is maximum, then back emf in solenoid is (u) a maximum. This occurs in AB part of the graph. So maximum back emf will be obtained between $5\; s

Since, the back emf at $t=3\; s$ is e.

Also, the rate of change of current at $t=3$,

and slope (s) of OA (from $t=0\; s$ to $t=5\; s$) $=\frac{1}{5}\frac{A}{s}$

So , we have

If $u=L\frac{1}{5}\left ( for\; t=3s,\frac{dI}{dt}=\frac{1}{5} \right )$.

where, L is a constant (coefficient of self -induction).

and emf is $\varepsilon =-L\frac{dI}{dt}$

Similarly, we have for others values.

For $5s

Thus, $at \; t=7 s, u_{1}=-3e$