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

ε 1 = − L 1 dI 2 dt − M 12 dI 1 dt

ε 1 = − L 1 dI 2 dt − M 12 dI 1 dt

ε 1 = − L 1 dI 1 dt - M 12 dI 1 dt

ε 1 = − L 1 dI 1 dt + M 12 dI 2 dt

ε 1 = − L 1 dI 2 dt − M 12 dI 1 dt

ε 1 = − L 1 dI 1 dt - M 12 dI 2 dt

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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

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$ε_{1} = - L_{1} \frac{{dI}_{1}}{dt} - M_{12} \frac{{dI}_{1}}{dt}$

$ε_{1} = - L_{1} \frac{{dI}_{1}}{dt} + M_{12} \frac{{dI}_{2}}{dt}$

$ε_{1} = - L_{1} \frac{{dI}_{2}}{dt} - M_{12} \frac{{dI}_{1}}{dt}$

$ε_{1} = - L_{1} \frac{{dI}_{1}}{dt} - M_{12} \frac{{dI}_{2}}{dt}$

answer is C.

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The total induced emf in coil 1 arises from both self-induction and mutual induction.

The self-induced emf in coil 1 due to its own changing current is given by:

The mutual inductance between the coils causes an additional induced emf in coil 1 due to the changing current in coil 2:

The net induced emf in coil 1 is the sum of both contributions:

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

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