Q68 / 75 · JEE Main 2025 (29 Jan Shift 1) · Physics

Consider \(\mathrm{I}_1\) and \(\mathrm{I}_2\) 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

  • A
    \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_2}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}\)
  • Correct answer
    B
    \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t}\)
  • C
    \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}\)
  • D
    \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}+\mathrm{M}_{12} \frac{\mathrm{dI}_2}{\mathrm{dt}}\)

Solution

$\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}}{1} \frac{\mathrm{I}_{12}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{~d} I_2}{\mathrm{dt}}\end{aligned}$

Hence, the correct answer is option: 'B'.

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