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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2025 · 29 January · Morning Shift · Physics · Electromagnetic Induction

Consider ( I _1) and ( I _2) are the currents flowing...

Consider (I\mathrm{I}_1) and (I\mathrm{I}_2) are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If (\mathrm{L}_1=) self inductance of coil (1, \mathrm{M}_{12}=) mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be Options

Options

  1. A(\varepsilon_1=-L_1 \frac{\mathrm{dI}_2}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t})
  2. B(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t})
  3. C(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t})
  4. D(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}+\mathrm{M}_{12} \frac{\mathrm{dI}_2}{\mathrm{dt}})

Correct answer

B. (\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t})

Step-by-step solution

ϕ1=L1I1+M12I2ε1=−dϕ1dt=−L1dI1I12dt−M12 dI2dt\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}}{1} \frac{\mathrm{I}_{12}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{~d} I_2}{\mathrm{dt}}\end{aligned}
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