Two Isolated Coils:
Let current i1 is alternating current and producing flux φ1 (per turn).
Then φ1 = φ11 + φ12
where φ 11 = flux linked with coil 1 only (leakage flux), and
φ 12 = flux reaches to coil 2 (mutual or useful flux).
Now the emf is induced in both the coils. In coil 1, emf e1 is induced because of its self inductance L1 and in coil 2 emf e2 is induced because of mutual inductance M between coils 1 and 2.
e1 = L1 di1 / dt (by Eq. (4))
Likewise, we may write the expression for e2
e2 = M (di1 /dt)
where M = mutual inductance among coils 1 and 2 (in henry)
e2 may also be written as :
e2 = N 2 d φ12 / dt
On equating Eqs. (6) and (7)
M di1/ dt = N2 (d φ12 /dt)
⇒ M = N2 (d φ12 /di1)
Assuming the zero initial conditions
M = N2 (φ12/ i2)
Now consider the reverse case, in which we give the supply to the secondary coil and finding the induced emf in primary coil.
Figure: Two Isolated Coils with Current i2
φ2 = φ21 + φ22
Where φ 21 = mutual flux,
φ 22 = leakage flux, and
φ 2 = total flux produced by i2.
So the induced emf
e2 = L2 ( di2 /dt)
(because of self inductance L2 of the coil).