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

Solution :
When the total flux associated with one coil links with the other i.e. a case of maximum flux linkage, then mutual induction in coil 1 due coil 2 is $${M_{12}} = \frac{{{N_2}{\phi _{{B_2}}}}}{{{i_1}}}$$
and mutual induction in coil 2 due to coil 1 is $${M_{21}} = \frac{{{N_1}{\phi _{{B_1}}}}}{{{i_2}}}$$
Similarly, self-inductance in coil 1 is $${L_1} = \frac{{{N_1}{\phi _{{B_1}}}}}{{{i_1}}}$$
and self-inductance in coil 2 is $${L_2} = \frac{{{N_2}{\phi _{{B_2}}}}}{{{i_2}}}$$
If all the flux of coil 2 links coil 1 and vice-versa, then $${\phi _{{B_2}}} = {\phi _{{B_1}}}$$
Since, $${M_{12}} = {M_{21}} = M,$$    hence we have
$$\eqalign{ & {M_{12}}{M_{21}} = {M^2} \cr & = \frac{{{N_1}{N_2}{\phi _{{B_1}}}{\phi _{{B_2}}}}}{{{i_1}{i_2}}} = {L_1}{L_2} \cr & \therefore {M_{\max }} = \sqrt {{L_1}{L_2}} \cr & {\text{Given,}}\,\,{L_1} = 2mH,{L_2} = 8mH \cr & \therefore {M_{\max }} = \sqrt {2 \times 8} = \sqrt {16} \cr & = 4\,mH \cr} $$