Mutual Inductance

A transformer is constructed of two coils placed so that the changing flux developed by one will link the other, as shown in Fig. 1.
Fig. 1: Defining the components of a transformer
This will result in an induced voltage across each coil. To distinguish between the coils, we will apply the transformer convention that
the coil to which the source is applied is called the primary, and the coil to which the load is applied is called the secondary.
For the primary of the transformer of Fig. 1, an application of Faraday's law will result in $$\bbox[10px,border:1px solid grey]{e_p = N_p { d \phi \over dt}} \,\, \text{(volts, V)} \tag{1}$$ revealing that the voltage induced across the primary is directly related to the number of turns in the primary and the rate of change of magnetic flux linking the primary coil. Or, $$\bbox[10px,border:1px solid grey]{e_p = L_p { di_p \over dt}} \,\, \text{(volts, V)} \tag{2}$$ revealing that the induced voltage across the primary is also directly related to the self-inductance of the primary and the rate of change of current through the primary winding.
The magnitude of $e_s$, the voltage induced across the secondary, is determined by $$\bbox[10px,border:1px solid grey]{e_s = N_s { d \phi_m \over dt}} \,\, \text{(volts, V)} \tag{3}$$ where $N_s$ is the number of turns in the secondary winding and $\phi_m$ is the portion of the primary flux $\phi_p$ that links the secondary winding.
If all of the flux linking the primary links the secondary, then $$\phi_m = \phi_p$$ and $$\bbox[10px,border:1px solid grey]{e_s = N_s { d \phi_p \over dt}} \,\, \text{(volts, V)} \tag{4}$$ The coefficient of coupling (k) between two coils is determined by $$k \, \text{(coefficient of coupling)} = { \phi_m \over \phi_p} \tag{5}$$
Since the maximum level of $\phi_m$ is $\phi_p$, the coefficient of coupling between two coils can never be greater than 1.
Fig. 2: Windings having different coefficients of coupling.
The coefficient of coupling between various coils is indicated in Fig. 2. In Fig. 2(a), the ferromagnetic steel core will ensure that most of the flux linking the primary will also link the secondary, establishing a coupling coefficient very close to 1. In Fig. 2(b), the fact that both coils are overlapping will result in the flux of one coil linking the other coil, with the result that the coefficient of coupling is again very close to 1. In Fig. 2(c), the absence of a ferromagnetic core will result in low levels of flux linkage between the coils. The closer the two coils are, the greater the flux linkage, and the higher the value of k, although it will never approach a level of 1. Those coils with low coefficients of coupling are said to be loosely coupled.
For the secondary, we have $$e_s = N_s { d \phi_m \over dt} = N_s { d k \phi_p \over dt}$$ and $$\bbox[10px,border:1px solid grey]{e_s = k N_s { d \phi_p \over dt}} \,\, \text{(volts, V)} \tag{6}$$ The mutual inductance between the two coils states that the
mutual inductance between two coils is proportional to the instantaneous change in flux linking one coil due to an instantaneous change in current through the other coil.
The mutual inductance between the two coils of Fig .1 is determined by $$\bbox[10px,border:1px solid grey]{M = N_s { d \phi_m \over di_p}} \,\, \text{(henries, H)} \tag{7}$$ or $$\bbox[10px,border:1px solid grey]{M = N_p { d \phi_p \over di_s}} \,\, \text{(henries, H)} \tag{8}$$ Note in the above equations that the symbol for mutual inductance is the capital letter M and that its unit of measurement, like that of self-inductance, is the henry.
In terms of the inductance of each coil and the coefficient of coupling, the mutual inductance is determined by $$\bbox[10px,border:1px solid grey]{M = k \sqrt{L_p L_s}} \, \text{(henries, H)} \tag{9}$$ The greater the coefficient of coupling (greater flux linkages), or the greater the inductance of either coil, the higher the mutual inductance between the coils. Relate this fact to the configurations of Fig. 2.
The secondary voltage es can also be found in terms of the mutual inductance if we rewrite Eq. (3) as $$e_s = N_s({d\phi_m \over di_p})({di_p \over dt})$$ and, since $M = Ns(d\phi_m/dip)$, it can also be written $$\bbox[10px,border:1px solid grey]{e_s = M{di_p \over dt}} \text{(volts, V)} \tag{10}$$ similarly, $$\bbox[10px,border:1px solid grey]{e_p = M{di_s \over dt}} \text{(volts, V)} \tag{11}$$
Example 1: For the transformer in Fig. 3:
a. Find the mutual inductance M.
b. Find the induced voltage $e_p$ if the flux $\phi_p$ changes at the rate of $450 mWb/s$.
c. Find the induced voltage $e_s$ for the same rate of change indicated in part (b).
d. Find the induced voltages $e_p$ and $e_s$ if the current $i_p$ changes at the rate of $0.2 A/ms$.
Fig. 3: Example 1.
Solution:
a. $$M = k \sqrt{L_p L_s} = 0.6 \sqrt{(200mH)(800mH)}\\ = 0.6 \sqrt{16 \times 10^{-2}} = 240mH$$ b. $$e_p = N_p{d \phi_p \over dt} \\ = (50)(450m Wb/s) = 22.5 V$$ c. $$e_s = kN_s{d \phi_p \over dt} \\ = (0.6)(100)(450m Wb/s) = 27 V$$ d. $$e_p = L_p{di_p \over dt} \\ = (200mH)(0.2 A/ms) = 40 V$$ $$e_s = M {di_p \over dt} \\ = (240mH)(0.2 A/ms) = 48 V$$