Q 12-06-045JEE MainJEE Main 2025 (29 Jan, Shift 1)Easy
Consider $I_1$ and $I_2$ are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If $L_1$ = self inductance of coil 1 and $M_{12}$ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
Answer: (B) $\varepsilon_1 = -L_1\dfrac{dI_1}{dt} - M_{12}\dfrac{dI_2}{dt}$
The flux through coil 1 comes from its own current and from coil 2's current: $\phi_1 = L_1I_1 + M_{12}I_2$.
$$\varepsilon_1 = -\frac{d\phi_1}{dt} = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$
Solution by Sreeraj P, M.Sc Physics