91Ó°ÊÓ

$$\begin{gathered} L_1=25\mathrm{mH}=0.025\mathrm{H} \\ {\mathrm{N}}_1=100 \\ {\mathrm{L}}_2=40\mathrm{mH}=0.040\mathrm{H} \\ {\mathrm{N}}_2=200 \\ \mathrm{M}=3.0\mathrm{mH}=0.003\mathrm{H} \\ \mathrm{I}=6.0\mathrm{mA}=0.006\mathrm{A} \\ \mathrm{di}/\mathrm{dt}=4.0\mathrm{A}/\mathrm{s} \end{gathered}$$

We have the formula for the magnetic flux induced in the coil, the self-induced emf and the mutually induced emf. So, using these equations and the given data, we can solve this problem.

Formula:

$$\begin{gathered} \mathrm{Ï•}=\frac{LI}{N} \\ \mathrm{Î\mu }=L\frac{di}{dt} \\ \mathrm{Ï•}=\frac{MI}{N} \\ \mathrm{Î\mu }=M\frac{di}{dt} \end{gathered}$$

We have,

$\mathrm{Ï•}=\frac{LI}{N}$

Magnetic flux ${\mathrm{Ï•}}_{12}$linking to coil 1 will be

$$\begin{gathered} {\mathrm{ϕ}}_{12}=\frac{L_{1}I}{N_1} \\ {\mathrm{ϕ}}_{12}=\frac{0.025×0.006}{100} \\ {\mathrm{ϕ}}_{12}=1.5×{10}^{-6}Wb \end{gathered}$$

We have the self – induced emf

$$\begin{gathered} \mathrm{Î\mu }=L\frac{di}{dt} \\ \mathrm{Î\mu }=0.025×4.0 \\ \mathrm{Î\mu }=0.1V \end{gathered}$$

The flux can be written in terms of the mutual inductance, the current and the number of turns.

$\mathrm{Ï•}=\frac{MI}{N}$

Magnetic flux ${\mathrm{Ï•}}_{21}$linking to coil 2, will be denoted as${\mathrm{Ï•}}_{21}$.

$$\begin{gathered} {\mathrm{ϕ}}_{21}=\frac{0.003×0.006}{200} \\ {\mathrm{ϕ}}_{21}=9×{10}^{-8}Wb \end{gathered}$$

EMF induced can be written in terms of the mutual inductance and the rate of change of the current.

$$\begin{gathered} \mathrm{Î\mu }=M\frac{di}{dt} \\ \mathrm{Î\mu }=0.003×4.0 \\ \mathrm{Î\mu }=0.012V \end{gathered}$$