Question

Consider \( I_1 \) and \( I_2 \) are the currents flowing simultaneously in two nearby coils 1 and 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. \( e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt} \)
• B. \( e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_1}{dt} \)
• 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:

Remember that the induced emf in a coil depends on both the self-inductance and the mutual inductance. The mutual inductance links the two coils and affects the induced emf due to current in both coils.

Answer 1

The Correct Option is B

The induced emf in a coil due to self-inductance and mutual inductance can be expressed using the following formula: \[ e_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt} \] However, since both currents are in the same direction and the mutual inductance influences the emf induced by both coils, the correct expression for the induced emf in coil 1, based on the mutual and self-inductance, is: \[ e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_1}{dt} \] This accounts for the self-induced emf in coil 1 and the influence of the mutual inductance between coils 1 and 2.