JEE Main 2025 Jan 29 Shift 1, Physics Q7: Faraday’s Law of Induction

JEE Main2025Jan 29, Shift 1Physics

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

Solution

Total flux through coil $1$ : $ϕ_{1} = L_{1} I_{1} + M_{12} I_{2}$ (self-flux plus the flux linked from coil $2$ ).

Faraday's law: $ε_{1} = - \frac{d ϕ _{1}}{d t} = - L_{1} \frac{d I _{1}}{d t} - M_{12} \frac{d I _{2}}{d t} .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q7: Faraday’s Law of Induction

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