Question

At a given instant the current and self-induced emf in an inductor are directed as indicated in Figure. (a) Is the current increasing or decreasing? (b) The induced emf is 17 V, and the rate of change of the current is 25 KA/s; find the inductance.

Short answer

1. Rate of change of the current,

$\frac{di}{dt}=25.0kA/s=25\times {10}^{3}A/s$

b. Induced emf, $\stackrel{,}{o}$=17.0 V

Step-by-step solution

Understanding the concept

We can use Faraday’s law of electromagnetic induction and the definition of inductance to find the inductance of the coil.

Formula:

$\stackrel{,}{o}=-\frac{d{\varphi }_B}{dt}$

${\varphi }_B=\mathrm{Li}$

(a) Find out if the is current increasing or decreasing 

At a given instant self-induced emf and current are directed towards the right of the page

So, the inductor will oppose the change in current, therefore current will be decreasing.

(b) Calculate the inductance

We have,

$\stackrel{,}{o}=-\frac{d{\varphi }_B}{dt}=-\frac{d}{dt}\left(Li\right)=-L\frac{di}{dt}$

Now taking the magnitude of above equation we get

$$\begin{gathered} L=\left(\frac{\stackrel{,}{o}}{{\displaystyle \frac{di}{dt}}}\right) \\ L=\frac{17.0}{25\times {10}^3} \\ L=0.68\times {10}^{-3}H=6.8\times {10}^{-4}H \end{gathered}$$