Question
Question: Current in a coil falls from \(2.5A\) to \(0.0A\) in \(0.1s\), inducing an emf of \[200V\]. Calculat...
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.
Solution
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=−LdtdI
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=−LdtdI
where
emf is the self-induced emf of a coil, when it encounters a changing magnetic flux
L is the value of self-inductance
dtdI is the change in the current flowing through the coil
Let this be equation 1.
Coming to our question, we are given that
emf=200VdtdI=0.1s0.0A−2.5A=−25As−1
Substituting these values in equation 1, we have
emf=−LdtdI⇒200V=−L×−25As−1⇒L=−25As−1−200V=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 As−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.