Question

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

Answer 1

Step 1: Find the time in which electromotive force is maximum.

Back electromotive force, is the electromotive force (voltage) that opposes the change in current which induced it.

So, when the rate of change of current is maximum, then back emf in solenoid u will be maximum.

According to the given graph current change is maximum in AB part.

So, maximum back emf will be obtained between
$5s<t<10s$.

Step 2: Find back emf at t = 7 s.

Formula used: back $emf=L\frac{dI}{dt}$

Given, back $emf$ at $t=3s=e$

And slope of $OA,=\frac{1}{5}A/s$

So, back $emf$ (at $t=3s$), $e=L\frac{dI}{dt}$

$e=L\left(\frac{1}{5}\right)$

At $t=7s,\frac{dI}{dt}=\frac{-3}{5}A/s$

So back $emf$, $u_1=-L\left(\frac{3}{5}\right)=-3e$

Step 3:Find back $emf$ at $t=15s$.

At $t=15s$, $\frac{dI}{dt}=\frac{1}{10}A/s$

So back $emf$, $u_2=L\left(\frac{1}{10}\right)$

$=\frac{e}{2}$

Step 4: Find back $emf$ at $t=40s$.

At $t=40s$, $\frac{dI}{dt}=0$

So back $emf$, $u_2=L\left(0\right)=0$

Final answer:
$5s<t<10s$;

$-3e;\frac{e}{2};0$.