Question

Consider two coils in which a current in one coil carrying \(6\) A causes the change in the flux in the second coil \(12\times 10^{-4}\) weber/turn. The second coil has \(2000\) turns. The mutual inductance between the coils is

• A. \(0.2\) H
• B. \(0.3\) H
• C. \(0.4\) H
• D. \(2.4\) H

Hint:

Mutual inductance is N times the flux per turn divided by the current.

Answer 1

The Correct Option is C

Step 1: Formula:
Flux linked with the second coil is \(N_2\phi_2\). Mutual inductance is \(M=\dfrac{N_2\phi_2}{I_1}\).

Step 2: Substitute:
\[ M=\frac{2000\times12\times10^{-4}}6=\frac{2.4}{6}=0.4\ \text{H} \]

Step 3: Check the Other Options:
0.2 H and 0.3 H come from using fewer turns, and 2.4 H is the total flux linkage without dividing by the current of 6 A.

Final Answer:
The mutual inductance is 0.4 H, option (C). \[ \boxed{\text{(C) } 0.4\ \text{H}} \]