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**Hint:**Let us first understand about emf. The electric potential produced by an electrochemical cell or by changing the magnetic field is known as electromotive force. Electromotive force is commonly abbreviated as EMF.

**Complete step by step answer:**

(a) Let us understand mutual inductance. This is the inductance set up between two coils with a connection between their fluxes. Henry is the SI unit of mutual inductance. The basic operating principle of the transformer, motors, generators, and any other electrical variable that interacts with another magnetic field is mutual inductance. The current flowing in one coil generates a voltage in an adjacent coil, which is known as mutual induction.

$L = \dfrac{{\phi (i)}}{i}$

$L = $Inductance, $\phi (i) = $Magnetic flux of current $i$ and $i = $Current.

Now let us come to the bit (a) of the question:

Using,

${E_1} = M\dfrac{{d{i_2}}}{{dt}}$...........................[$Equation - 1$]

Where, ${E_1} = $emf in the 1st coil, $M = $Mutual inductance and $\dfrac{{d{i_2}}}{{dt}} = $rate of current increase in coil 2.

(b) Given, ${E_1} = 25mV = \dfrac{{25}}{{1000}}V = 0.025\,V$

$\dfrac{{d{i_2}}}{{dt}} = 15\,A/s$

Substitute the above values into the $Equation - 1$

$0.025 = 15M \\

\Rightarrow M = \dfrac{{0.025}}{{15}} \\

\Rightarrow M = 1.66 \times {10^{ - 3}}H \\$

Therefore the Mutual Inductance is $M = 1.66 \times {10^{ - 3}}\,H$

Now, let us solve the bit (b) of the problem: as we know,

$|M| = \dfrac{{e_1}\,dt}{di_2}= \dfrac{{{\phi _2}}}{{{i_1}}}$

$\Rightarrow {\phi _2} =\dfrac{{e_1}\,dt\,{i_1}}{di_2} \\

\Rightarrow {\phi _2} = \dfrac{{(25.0 \times {{10}^{ - 3}})(3.60)}}{{15}} \\

\Rightarrow {\phi _2} = 6 \times {10^{ - 3}} \\

\therefore {\phi _2} = 6\,mWb$

**Hence, the flux linkage in coil $2$ is $6\,mWb$.**

**Note:**Negative electromotive force is possible. Consider the case of an inductor that generates an EMF that is in opposition to the incoming force. The generated EMF is then interpreted as negative because the flow direction is opposite that of the real force. As a result, the electromotive force is possible to be negative.

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