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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…

JEE Main 2025 — Physics Electromagnetism

2025mcqmedium

Consider I1\mathrm{I}_1I1​ and I2\mathrm{I}_2I2​ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1=\mathrm{L}_1=L1​= self inductance of coil 1,M12=1, \mathrm{M}_{12}=1,M12​= mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be Options

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    ε1=−L1dI2dt−M12dI1dt\varepsilon_1=-L_1 \frac{\mathrm{dI}_2}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}ε1​=−L1​dtdI2​​−M12​dtdI1​​

  2. B

    ε1=−L1dI1dt−M12dI2dt\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t}ε1​=−L1​dtdI1​​−M12​dtdI2​​

  3. C

    ε1=−L1dI1dt−M12dI1dt\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}ε1​=−L1​dtdI1​​−M12​dtdI1​​

  4. D

    ε1=−L1dI1dt+M12dI2dt\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}+\mathrm{M}_{12} \frac{\mathrm{dI}_2}{\mathrm{dt}}ε1​=−L1​dtdI1​​+M12​dtdI2​​

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