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Physics · Electromagnetic Induction

JEE Main 2025 — 29 January, Morning Shift — Question 29

Consider I1I_{1} and I2I_{2} are the currents flowing simultaneously in two nearby coils

1&21 \& 2, respectively. If L1=L_{1}= self inductance of coil 1 , M12=\mathrm{M}_{12}=

mutual inductance of coil 1 with respect to coil 2 , then the value of induced emf in coil 1 will be

  1. Option A:

    ε1=−L1dIdt+M12dII2dt\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}}{\mathrm{dt}}+\mathrm{M}_{12} \frac{\mathrm{dI} I_{2}}{\mathrm{dt}}

  2. Option B:

    ε1=−L1dI1dt−M12dI1dt\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}

  3. Option C:

    ε1=−L1dI1dt−M12dII2dt\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI} I_{2}}{\mathrm{dt}}

    Correct
  4. Option D:

    ε1=−L1dI2dt−M12dI1dt\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}

Answer: C

Step-by-step solution

ϕ1=L1I1+M12I2\phi_{1}=L_{1} I_{1}+M_{12} I_{2}

ε1=−dϕ1dt=−L1dI1dt−M12dI2dt\varepsilon_{1}=-\frac{\mathrm{d} \phi_{1}}{\mathrm{dt}}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Self-Inductance and Mutual Inductance and Energy Density

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