91Ó°ÊÓ

1. Inductance for first coil =$L_1$
2. Inductance for second coil =$L_2$
3. Mutual Inductance for both coils =$M$
4. Current flowing through the coils = i

We have the formula for the self-inductance and mutual inductance for the coil. If we find the equivalent inductance for the situation we can get the required answer.

We can reverse engineer the given equation to find the connection between the given two coils to yield an equivalent inductance of ${\mathit{L}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{L}}_{\mathbf{1}}\mathbf{+}{\mathit{L}}_{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{M}$

Formula:

$$\begin{gathered} \mathrm{Î\mu }=-Ldi/dt \\ \mathrm{Î\mu }=-Mdi/dt \end{gathered}$$

Self-induction for the first coil,

${\mathrm{Î\mu }}_{S1}=-L_1\frac{di}{dt}$

Mutual inductance for the first coil,

${\mathrm{Î\mu }}_{S1}=-M_{1}di/dt$

From this the net inductance for first coil will be,

$$\begin{gathered} {\mathrm{Î\mu }}_1={\mathrm{Î\mu }}_{S1}+{\mathrm{Î\mu }}_{M1} \\ {\mathrm{Î\mu }}_{S1}=-\left(L_1+M\right)di/dt \end{gathered}$$

Self-induction for the second coil,

${\mathrm{Î\mu }}_{\mathrm{S}2}={\mathrm{L}}_2\frac{\mathrm{di}}{\mathrm{dt}}$

Mutual inductance for the second coil,

${\mathrm{Î\mu }}_{\mathrm{M}2}=-Mdi/dt$

From this the net inductance for second coil will be,

$$\begin{gathered} {\mathrm{Î\mu }}_2={\mathrm{Î\mu }}_{S2}+{\mathrm{Î\mu }}_{\mathrm{M}2} \\ {\mathrm{Î\mu }}_2=-\left({\mathrm{L}}_2+\mathrm{M}\right)\mathrm{di}/\mathrm{dt} \end{gathered}$$

The equivalent inductance for both coils will be,

$$\begin{gathered} L_{eq}={\mathrm{Î\mu }}_1={\mathrm{Î\mu }}_2 \\ {L}_{eq}=\left(-\left(L_1+M\right)\frac{di}{dt}\right)+\left(-\left(L_2+M\right)\frac{di}{dt}\right) \\ {L}_{eq}=-\left(L_1+L_2+2M\right)\frac{di}{dt} \end{gathered}$$

So, this will be the emf which would be produced if both the coils are replaced by the single coil

Now, imagine we reverse the current flowing through the coil 2, i.e. it will enter from the back of the coil and will exit from the front of the coil.

This will result into

Self-induction for the first coil,

${\mathrm{Î\mu }}_{S1}=-L_1\frac{di}{dt}$

Mutual inductance for the first coil,

${\mathrm{Î\mu }}_{M1}=-M_{1}di/dt$

From this the net inductance for first coil will be,

$$\begin{gathered} {\mathrm{Î\mu }}_1={\mathrm{Î\mu }}_{S1}+{\mathrm{Î\mu }}_{M1} \\ {\mathrm{Î\mu }}_1=-\left(L_1-M\right)di/dt \end{gathered}$$

Self-induction for the second coil,

${\mathrm{Î\mu }}_{\mathrm{S}2}=-{\mathrm{L}}_2\frac{\mathrm{di}}{\mathrm{dt}}$

Mutual inductance for the second coil,

role="math" localid="1661428555396" ${\mathrm{Î\mu }}_{\mathrm{M}2}=Mdi/dt$

From this the net inductance for second coil will be,

$$\begin{gathered} {\mathrm{Î\mu }}_2={\mathrm{Î\mu }}_{S2}+{\mathrm{Î\mu }}_{\mathrm{M}2} \\ {\mathrm{Î\mu }}_2=-\left({\mathrm{L}}_2-\mathrm{M}\right)\mathrm{di}/\mathrm{dt} \end{gathered}$$

The equivalent inductance for both coils will be,

$$\begin{gathered} L_{eq}={\mathrm{Î\mu }}_1+{\mathrm{Î\mu }}_2 \\ {L}_{eq}=\left(-\left(L_1-M\right)\frac{di}{dt}\right)+\left(-\left(L_2-M\right)\frac{di}{dt}\right) \\ {L}_{eq}=-\left(L_1+L_2-2M\right)\frac{di}{dt} \end{gathered}$$

So this will be the emf which would be produced if both the coils are replaced by the single coil.

So we have to connect the coil 2 in reverse direction.