Question

Two coils P and Q are kept near each other. When no current flows through coil P and current in coil Q increases at the rate of 10 A/s, the e.m.f. in coil P is 12 mV. When coil Q carries no current and current 1.5 A flows through coil P, the magnetic flux linked with the coil Q is

• A. \(1.8\text{ mWb}\)
• B. \(1.5\text{ mWb}\)
• C. \(1.2\text{ mWb}\)
• D. \(0.9\text{ mWb}\)

Hint:

Find the mutual inductance from the first case and use the fact that it is the same in both directions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The emf in one coil is \(e=M\dfrac{dI}{dt}\) due to the other, and the flux is \(\phi=MI\). The mutual inductance \(M\) is the same for both coils.

Step 2: Find M
\(M=\dfrac{e_P}{dI_Q/dt}=\dfrac{12\times10^{-3}}{10}=1.2\times10^{-3}\text{ H}\)

Step 3: Flux in Q
\(\phi_Q=MI_P=1.2\times10^{-3}\times1.5\)

Step 4: Result
\(\phi_Q=1.8\times10^{-3}\text{ Wb}=1.8\text{ mWb}\)

Final Answer:
The flux linked with coil Q is \(1.8\text{ mWb}\). \[ \boxed{\text{(A) }1.8\text{ mWb}} \]