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 determine the value of the induced electromotive force (emf) in coil 1 when currents are flowing in two nearby coils, we need to consider both self-induction and mutual induction effects. The self-induced emf in coil 1, due to its own current, is given by:

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

where:

• \(L_1\) is the self-inductance of coil 1.
• \(\frac{dI_1}{dt}\) is the rate of change of current in coil 1.

Additionally, the mutual induced emf in coil 1, due to the current in coil 2, is expressed as:

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

where:

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

The total induced emf in coil 1, considering both self-induction and mutual induction, is:

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

Substituting the expressions for \(e_{self}\) and \(e_{mutual}\), we get:

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

Answer 2

Step 1: Understand the problem.
We are considering two nearby coils carrying currents \( I_1 \) and \( I_2 \), respectively. Due to the time-varying currents, each coil produces an induced emf caused by both self-induction and mutual induction effects.

Step 2: Self-induced emf in coil 1.
When the current in coil 1 changes with time, it induces an emf in itself according to Faraday’s law of electromagnetic induction. The self-induced emf is given by:
\[ e_{\text{self}} = -L_1 \frac{dI_1}{dt}, \] where \( L_1 \) is the self-inductance of coil 1, and the negative sign indicates that the induced emf opposes the change in current (Lenz’s law).

Step 3: Mutually induced emf in coil 1 due to coil 2.
When the current in the nearby coil 2 changes, it induces an emf in coil 1 because of mutual inductance between the coils. The mutually induced emf in coil 1 is given by:
\[ e_{\text{mutual}} = M_{12} \frac{dI_2}{dt}, \] where \( M_{12} \) is the mutual inductance of coil 1 with respect to coil 2.

Step 4: Total induced emf in coil 1.
The total emf in coil 1 is the sum of both self-induced and mutually induced emf. Hence,
\[ e_1 = e_{\text{self}} + e_{\text{mutual}}. \] Substituting the values from above:
\[ e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}. \]

Step 5: Final Expression.
Therefore, the induced emf in coil 1 due to both self and mutual induction is:
\[ \boxed{e_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}}. \]

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