Question

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

• A. \( e_1 = -L_1 \frac{dI_2}{dt} + M_{12} \frac{dI_1}{dt} \)
• B. \( e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt} \)
• C. \( e_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt} \)
• D. \( e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_1}{dt} \)

Hint:

To find the induced emf in a coil with mutual induction, use the principle of mutual inductance in combination with the self-inductance and rate of change of current in both coils.

Answer 1

The Correct Option is B

The induced EMF in a coil is given by Faraday’s law of induction: \[ e_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt} \] where \( L_1 \frac{dI_1}{dt} \) is the self-induced emf due to coil 1's current and \( M_{12} \frac{dI_2}{dt} \) is the mutual induced emf due to the changing current in coil 2. Thus, the correct equation for the induced EMF in coil 1 is: \[ e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt} \] Thus, the correct answer is (2).