JEE Physics 2025 Question with Solution

Given: Two nearby coils carry currents $I_1$ and $I_2$ simultaneously. For coil $1$, self inductance is $L_1$ and mutual inductance with respect to coil $2$ is $M_{12}$.

Find: The induced emf in coil $1$.

The induced emf in coil $1$ has two contributions: self-induction due to change in $I_1$ and mutual induction due to change in $I_2$.

For self-induction,

$$e_{\text{self}} = -L_1 \frac{dI_1}{dt}$$

For mutual induction,

$$e_{\text{mutual}} = M_{12} \frac{dI_2}{dt}$$

Therefore, the total induced emf in coil $1$ is

$$e_1 = e_{\text{self}} + e_{\text{mutual}}$$

Substituting,

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

Therefore, the correct option is B.

Given: Two nearby coils carry time-varying currents $I_1$ and $I_2$.

Find: Expression for induced emf in coil $1$.

1. A changing current in coil $1$ produces self-induced emf in the same coil.

$$e_{\text{self}} = -L_1 \frac{dI_1}{dt}$$

The negative sign follows Lenz's law.

2. A changing current in nearby coil $2$ produces mutually induced emf in coil $1$.

$$e_{\text{mutual}} = M_{12} \frac{dI_2}{dt}$$

3. Total induced emf in coil $1$ is the sum of these two effects.

$$e_1 = e_{\text{self}} + e_{\text{mutual}}$$

4. Substituting the expressions,

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

Therefore, the induced emf in coil $1$ is $e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}$, so the correct option is B.