Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

Given,

$L_1=2mH, L_2=8mH$

Total flux associated with the coil that is linked with other mutual inductance,

$$\begin{gathered} M_{12}=\frac{N_2{\varphi }_B}{i_1} \\ {M}_{21}=\frac{N_1{\varphi }_B}{i_2} \end{gathered}$$

similarly self-inductance, $L_1=\frac{N_1{\varphi }_{B_1} }{i_1}$and $L_2=\frac{N_2{\varphi }_{B_2} }{i_2}$ 

If all flux of the coil 2 is linked with coil 1 , vice-versa

${\varphi }_{B_2}={\varphi }_{B_1}$

We know that $$\begin{gathered} M_{12}=M_{21}=M \\ \Rightarrow M_{12}=M_{21}=M^2 \\ \Rightarrow \frac{N_1{N}_2{\varphi }_{B_1}{\varphi }_{B_2}}{i_1{i}_2}=L_1{L}_2 \\ \therefore M_{max}=\sqrt{L_1{L}_2} \end{gathered}$$

substituting the value of $L_{1 }and L_2$

$$\begin{gathered} \therefore M_{max}=\sqrt{L_1{L}_2} \\ \therefore M_{max}=\sqrt{2\times 8} \\ =\sqrt{16} \\ =4mH \end{gathered}$$

Hence the mutual inductance between the coils is $4mH$.