Question

When magnetic flux changes from $6.5 \times 10^{-2}$ Wb to $11 \times 10^{-2}$ Wb and the change in current is 0.03 A, the coefficient of mutual inductance will be ______.

• A. 1.0 H
• B. 1.2 H
• C. 1.5 H
• D. 1.8 H

Hint:

Mutual inductance ($M$) is simply the ratio of "how much flux is produced" to "how much current caused it". Just divide the net change in flux by the net change in current!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a change in magnetic flux in one coil caused by a change in current in a neighboring coil. We must calculate the coefficient of mutual inductance ($M$) linking the two coils.

Step 2: Detailed Explanation:
By the fundamental definition of mutual inductance, the total magnetic flux ($\Phi$) linked with the secondary coil is directly proportional to the current ($I$) flowing through the primary coil:
$\Phi = M \cdot I$
For any macroscopic change in current, there is a corresponding direct change in flux:
$\Delta \Phi = M \cdot \Delta I$
We can rearrange this formula to solve for the coefficient of mutual inductance ($M$):
$M = \frac{\Delta \Phi}{\Delta I}$
Let's calculate the change in flux ($\Delta \Phi$):
Initial flux ($\Phi_1$) = $6.5 \times 10^{-2} \text{ Wb}$
Final flux ($\Phi_2$) = $11 \times 10^{-2} \text{ Wb}$
$\Delta \Phi = \Phi_2 - \Phi_1 = (11 - 6.5) \times 10^{-2} = 4.5 \times 10^{-2} \text{ Wb}$
We are given the change in current ($\Delta I$):
$\Delta I = 0.03 \text{ A} = 3 \times 10^{-2} \text{ A}$
Substitute these values into the mutual inductance formula:
$M = \frac{4.5 \times 10^{-2}}{3 \times 10^{-2}}$
The $10^{-2}$ terms cancel out perfectly:
$M = \frac{4.5}{3} = 1.5 \text{ H}$

Step 3: Final Answer:
The coefficient of mutual inductance is 1.5 H, matching option (c).