Two coils have self inductance \(L_{1}\) and \(L_{2}\) and mutual inductance between them is M. Derive a mathematical expression for co – efficient of coupling (k) for these coils. Find an expression for the energy stored in a magnetic field.
Two coils are said to be mutually coupled by magnetic flux if a part or whole of the flux generated by one coil is linked with the other coil. Let us again take two coils having \(N_1\) and \(N_2\) as the number of turns. If \(I_1\) and \(I_2\) be the currents flowing in coil 1 and 2 respectively having the same reluctance R then the fluxes linked with coils 1 and 2 are \(\mathrm{\Phi}_1=\frac{N_1I_1}{R}, \mathrm{\Phi}_2=\frac{N_2I_2}{R}\) …(1)
If a fraction k of the total flux produced by current \(I_1\) in coil 1 is liked with coil 2 then \(\mathrm{\Phi}_{12}=k\mathrm{\Phi}_1\ and\ \mathrm{\Phi}_{21}=k\mathrm{\Phi}_2\) …(2)
The mutual inductance between two coils are
\(M_{12}=N_2\frac{\mathrm{\Phi}_{12}}{I_1}\) …(3)
and \(M_{21}=N_1\frac{\mathrm{\Phi}_{21}}{I_2}\) …(4)
Substituting the value of \(\mathrm{\Phi}_{12}\) and \(\mathrm{\Phi}_{21}\) from equation (1) into equation (3) and (4), we have
\(M_{12}=k\frac{N_1N_2}{R}\) …(5)
and \(M_{21}=k\frac{N_1N_2}{R}\) …(6)
It is assuming that the circuit is bilateral, we have
\(M_{12}=M_{21}=M\)
\(M^2=M_{12}\times M_{21}=k\frac{N_1N_2}{R}\times k\frac{N_1N_2}{R}\)
\(M^2=\left(k\frac{N_1N_2}{R}\right)^2\) …(7)
The self inductance of coils 1 and 2 are
\(L_1=\frac{{N_1}^2}{R} and L_2=\frac{{N_2}^2}{R}\) …(8)
Using equation (7) and (8), we have
\(M^2=k^2L_1L_2\)
Therefore the coupling coefficient \(k=M/\sqrt{L_1L_2}\)
The above inductor having inductance L is connected to the bettery through key K. In this case inducted e.m.f is given by
\(e=-L\frac{dL}{dt}\) …(1)
To drive the current through the inductor against the induced e.m.f ‘e’, the external voltage is applied from the battery has e.m.f E and in this case E=-e
Therefore \(E=L\frac{dL}{dt}\)
Let an infinitesimal charge dq be driven through the inductor. So the work done by the external voltage to do this is given by:
dw=Edq
Or \(dw=L\frac{dL}{dt}\times dq=Ldi(\frac{dq}{dt})\), we know that \(i=\frac{dq}{dt}\)
Hence \(dw=L\times i\times di\)
Total work done to maintain the maximum value of current \(I_0\) through the inductor is given by
\(\int d w=\int_{0}^{I_0}{L\times i\ di}\)
Or \(W=L\left[\frac{\left(I_0\right)^2}{2}\right]\)
The work done in increasing the current flowing through the inductor is stored as the potential energy in the magnetic field.
Hence potential energy \(E_p=\frac{1}{2}L\left(I_0\right)^2\)
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