Verified Solution • 2025–26 Session moderate

QJEEM25Pha8bf: JEE Main 2025 Physics - Class 12 JEEM25Pha8bf Solution: Consider I 1 and I 2 are the currents flowing simultaneously in …

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 :

Multiple Choice Options

  • A e1 = -L1 dI2dt\frac{dI_2}{dt} - M12 dI1dt\frac{dI_1}{dt}
  • B e1 = -L1 dI1dt\frac{dI_1}{dt} + M12 dI2dt\frac{dI_2}{dt}
  • C e1 = -L1 dI1dt\frac{dI_1}{dt} - M12 dI1dt\frac{dI_1}{dt}
  • D e1 = -L1 dI1dt\frac{dI_1}{dt} - M12 dI2dt\frac{dI_2}{dt}

Step-by-Step Verified Solution

ϕ1=L1I1+M12I2ε1=−dϕ1dt=−L1dI1dt−M12dIdt

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