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

To ascertain the induced electromotive force (emf) in coil 1 when currents flow in two adjacent coils, both self-induction and mutual induction effects must be accounted for. The emf self-induced in coil 1 by its own current is defined as:

\(e_{self} = -L_1 \frac{dI_1}{dt}\)

In this equation:

• \(L_1\) represents the self-inductance of coil 1.
• \(\frac{dI_1}{dt}\) denotes the rate at which the current in coil 1 changes.

Furthermore, the emf mutually induced in coil 1, resulting from the current in coil 2, is expressed by:

\(e_{mutual} = M_{12} \frac{dI_2}{dt}\)

Where:

• \(M_{12}\) signifies the mutual inductance between coil 1 and coil 2.
• \(\frac{dI_2}{dt}\) is the rate of change of current in coil 2.

The aggregate induced emf in coil 1, encompassing both self-induction and mutual induction, is calculated as:

\(e_1 = e_{self} + e_{mutual}\)

By substituting the previously defined expressions for \(e_{self}\) and \(e_{mutual}\), the total induced emf in coil 1 is:

\(e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}\)