1. Flux linkage in coil 1

When two nearby coils carry currents simultaneously, the magnetic flux linked with coil 1 has two contributions:

• due to its own current $I_1$
• due to the current in coil 2 $I_2$

So, total flux linkage of coil 1 is

$\lambda_1 = L_1 I_1 + M_{12} I_2$

where:

• $L_1$ = self inductance of coil 1
• $M_{12}$ = mutual inductance contribution in coil 1 due to coil 2

2. Use Faraday's law

Induced emf in coil 1 is

$e_1 = -\frac{d\lambda_1}{dt}$

Substitute $\lambda_1$:

$e_1 = -\frac{d}{dt}(L_1 I_1 + M_{12} I_2)$

Assuming $L_1$ and $M_{12}$ are constants,

$e_1 = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$

3. Match with the options

• A: $-L_1 \dfrac{dI_2}{dt} - M_{12} \dfrac{dI_1}{dt}$ → incorrect, derivatives are interchanged.
• B: $-L_1 \dfrac{dI_1}{dt} + M_{12} \dfrac{dI_2}{dt}$ → incorrect sign of mutual term.
• C: $-L_1 \dfrac{dI_1}{dt} - M_{12} \dfrac{dI_1}{dt}$ → incorrect, second term should involve $\dfrac{dI_2}{dt}$.
• D: $-L_1 \dfrac{dI_1}{dt} - M_{12} \dfrac{dI_2}{dt}$ → correct.

4. Final answer

$\boxed{e_1 = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}}$

Hence, Option D is correct.