Question

A pair of adjacent coils has a mutual inductance of $1.5\,H$. If the current in one coil changes from $0$ to $20\,A$ in $0.5$ seconds, what is the change of the flux linkage with the other coil?

Answer 1

Hint:We shall use the concept of induced emf here. Any varying current in a magnetic coupled coil would produce an induced emf in the other coil which is given by $e = M\dfrac{{di}}{{dt}}$ where $e$ is the induced emf and $M$ is the mutual inductance. Also, the varying current is accompanied with a change in flux linkage. This change can be expressed in the form of induced emf as \[e = \dfrac{{d\phi }}{{dt}}\] where $\phi $ is the flux linkage of two coils. We shall equate both the equations to obtain a relation between the flux linkage, mutual inductance and the change in current. Making proper substitutions, we will get our answer.

Complete step by step answer:
A varying current produces an emf in the coil magnitude of which is given by $e = M\dfrac{{di}}{{dt}}$ where e is the induced emf and M is the mutual inductance.
This induced emf can also be expressed as the change in the flux linkage as \[e = \dfrac{{d\phi }}{{dt}}\] where $\phi $ is the flux linkage of two coils.
Hence, we can say that \[e = \dfrac{{d\phi }}{{dt}} = M\dfrac{{di}}{{dt}}\]
This can be rewritten as
\[d\phi = Mdi\,\,\,\,\,\,\,\,\,\,\,\,.........(1)\]
The change in current, $di = (20 - 0)\,A$
$ \Rightarrow di = 20\,A$
The time interval in which the current changes is $dt = 0.5\,s$
The mutual inductance of the coils is given to be $M = 1.5\,H$
Substituting in the equation 1 we get,
\[d\phi = 1.5 \times 20\]
\[ \therefore d\phi = 30\,Wb\]

Hence,the change of the flux linkage with the other coil is $30\,Wb$.

Note:Here we have focused only on the magnitude of the induced emf. Its direction is given by Lenz’s law which states that the induced current and thus the induced emf is set up in such a way that it opposes the cause of change. Here we cannot specify a particular direction for the emf produced since the direction was not given in the question. Thus, we calculated only the magnitude of the emf and made further simplifications.