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

(C) $\varepsilon_{1}=-L_{1} \frac{d I_{1}}{d t}-M_{12} \frac{d I_{2}}{d t}$

(A) $\varepsilon_{1}=-L_{1} \frac{d I_{1}}{d t}+M_{12} \frac{d I_{2}}{d t}$

(B) $\varepsilon_{1}=-L_{1} \frac{d I_{1}}{d t}-M_{12} \frac{d I_{1}}{d t}$

(D) $\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}$

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Electromagnetic Induction

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

(A) $ε_{1} = - L_{1} \frac{d I _{1}}{d t} + M_{12} \frac{d I _{2}}{d t}$
(B) $ε_{1} = - L_{1} \frac{d I _{1}}{d t} - M_{12} \frac{d I _{1}}{d t}$
(C) $ε_{1} = - L_{1} \frac{d I _{1}}{d t} - M_{12} \frac{d I _{2}}{d t}$ ✓ Correct
(D) $ε_{1} = - L_{1} \frac{dI _{2}}{dt} - M_{12} \frac{dI _{1}}{dt}$

Answer: (C)

Coil 1 links flux from its own current and from the current in coil 2:

ϕ_{1} = L_{1} I_{1} + M_{12} I_{2}

By Faraday's law, the induced emf in coil 1 is

ε_{1} = - \frac{d ϕ _{1}}{d t} = - L_{1} \frac{d I _{1}}{d t} - M_{12} \frac{d I _{2}}{d t}

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option