JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 12)
Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :
e1 = -L1$\frac{dI_2}{dt}$ - M12$\frac{dI_1}{dt}$
e1 = -L1$\frac{dI_1}{dt}$ + M12$\frac{dI_2}{dt}$
e1 = -L1$\frac{dI_1}{dt}$ - M12$\frac{dI_1}{dt}$
e1 = -L1$\frac{dI_1}{dt}$ - M12$\frac{dI_2}{dt}$
Explanation
$$\begin{aligned}
& \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\
& \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}_1}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}}{\mathrm{dt}}
\end{aligned}$$
Comments (0)
