Question

Current in a coil falls from $2.5A$ to $0.0A$ in $0.1s$, inducing an emf of \[200V\]. Calculate the value of self-inductance.

Answer 1

Hint: Self-induction in a coil is the process by which the coil induces a current inside the coil, when it encounters a changing magnetic flux. The value of self-inductance of the coil is directly proportional to the induced $emf$ in the coil. At the same time, it is inversely proportional to the rate of change of current in the coil.

Formula used: $emf=-L\dfrac{dI}{dt}$

Complete step by step answer:
Self-induction in a coil is the process by which the coil induces a current inside the coil, when it encounters a changing magnetic flux. The value of self-inductance of the coil is directly proportional to the induced $emf$ in the coil. At the same time, it is inversely proportional to the rate of change of current in the coil. Mathematically, $emf$ induced in the coil is given by
$emf=-L\dfrac{dI}{dt}$
where
$emf$ is the self-induced emf of a coil, when it encounters a changing magnetic flux
$L$ is the value of self-inductance
$\dfrac{dI}{dt}$ is the change in the current flowing through the coil
Let this be equation 1.
Coming to our question, we are given that
$\begin{align}
  & emf=200V \\
 & \dfrac{dI}{dt}=\dfrac{0.0A-2.5A}{0.1s}=-25A{{s}^{-1}} \\
\end{align}$
Substituting these values in equation 1, we have
$emf=-L\dfrac{dI}{dt}\Rightarrow 200V=-L\times -25A{{s}^{-1}}\Rightarrow L=\dfrac{-200V}{-25A{{s}^{-1}}}=8H$
Let this be equation 2.

Therefore, from equation 2, we can conclude that the value of self-inductance in the given case is equal to $8H$.

Note: The SI unit of self-inductance is $henry(H)$. It is equivalent to the ratio of $V$ to $A{{s}^{-1}}$, as can be understood from the final equation. Students need not get confused with the negative sign in the formula used. This suggests that the self-induced $emf$ of the coil tends to restrict further changes in flux, inside the coil.