Consider $$I_{1}$$ and $$I_{2}$$ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If $$L_{1}$$ = self inductance of coil 1, $$M_{12}$$ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be Options

We need to find the EMF induced in coil 1 when it carries current $$I_1$$ and a nearby coil 2 carries current $$I_2$$.

The total flux through coil 1 has two contributions:

(i) Self-flux due to its own current: $$\Phi_{\text{self}} = L_1 I_1$$

(ii) Mutual flux due to current in coil 2: $$\Phi_{\text{mutual}} = M_{12} I_2$$

The total flux linkage with coil 1:
$$\Phi_1 = L_1 I_1 + M_{12} I_2$$

By Faraday's law, the EMF induced in coil 1 is:
$$\varepsilon_1 = -\frac{d\Phi_1}{dt} = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$

The correct answer is Option (2): $$\varepsilon_1 = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$.