Question

An inductor of $5$ Henry inductance carries the current of $2\,A$. How can a $50\,volt$ self inductance be made to appear at the end of the inductor ?

Answer 1

Hint:EMF is the electromotive force which is produced by either electrochemical cell or by changing the magnetic field. The emf is induced in a coil due to the change of its own flux linked with it is known as Self-induced emf. Emf is induced without physical motion or flux. It is usually denoted by the letter $\varepsilon $.

Complete step by step answer:
In the above problem, it is given that the inductance of the inductor, $L = 5H$ and it carries a current which is equal to $2A$.
Induced emf= $50V$
By Using Faraday’s law, we know that the expression for inductance is given by
$_{\varepsilon = N\dfrac{{d\phi }}{{dt}}}$
From our prior knowledge we know that $\phi = \dfrac{{{\mu _0}NIA}}{l}$ . Now substituting this value in the above equation we get,
$\varepsilon = \dfrac{d}{{dt}}\left( {\dfrac{{{\mu _0}{N^2}IA}}{l}} \right) \\
\Rightarrow \varepsilon = \dfrac{{{\mu _0}{N^2}A}}{l}\dfrac{{dI}}{{dt}}$
$\Rightarrow \varepsilon = L\dfrac{{dI}}{{dt}}$
Now substitute all the values in the above equation to get the final solution,
$L\dfrac{{di}}{{dt}} = 50 \\
\Rightarrow dt = \dfrac{{L \times di}}{{50}} $
Now substitute the values of $di$ and $L$ in the above equation we get
$ \Rightarrow dt = \dfrac{{5 \times 2}}{{20}}$
$ \therefore dt = 0.2s$

Hence, the induced emf can be produced by decreasing the current to zero at 0.2 s.

Note: Inductance is the ability of an inductor to store energy and it does this in the magnetic field that is created by the flow of electrical current . As a result of the magnetic field associated with the current flow, the inductor generates an opposing voltage proportional to the rate of change in the current circuit.