Q.A pair of adjacent coils has a mutual inductance of $1.5\ \text{H}$. If the current in one coil changes from $0$ to $20\ \text{A}$ in $0.5\ \text{s}$, what is the change of flux linkage with the other coil?

The key idea is Mutual Inductance: the flux linkage in one coil due to current in another is $\Phi_2 = M I_1$, so a change in current causes a proportional change in flux linkage.

Step 1: Write the relation for flux linkage.

For coil 2 due to current $I_1$ in coil 1:

$\Phi_2 = M I_1$.

Step 2: The change in flux linkage is

$\Delta \Phi_2 = M \, \Delta I_1$.

Step 3: Substitute values: …

Mutual inductance is a measure of how effectively a changing current in one coil induces a magnetic flux through another coil. The key idea is that the flux linkage in the second coil is proportional to the current in the first coil, with the constant of proportionality being the mutual inductance $M$.

The definition is:

If a current $I_1$ flows in coil 1, the flux linkage (total magnetic flux linking all turns) in coil 2 is

where $\Phi_2$ is in weber-turns (Wb-turns) and $M$ is in henries (H). This is a direct, linear relationship — no time derivative involved yet.

When the current changes, the flux linkage changes by the same proportion. So the change in flux linkage $\Delta \Phi_2$ is simply $M$ times the change in current $\Delta I_1$:

This is the cleanest way to get the answer. The time interval (0.5 s) is irrelevant for the change in flux linkage — it only matters if you were asked for the induced emf (which would be $M \frac{\Delta I}{\Delta t}$).

Now let’s plug in the numbers.

1. Identify the given values Mutual inductance: $M = 1.5\ \text{H}$ …

Method: Faraday's Law of Electromagnetic Induction (using Mutual Inductance)

This problem is solved using the definition of mutual inductance — which directly relates the induced flux linkage in one coil to the current change in the other coil.

Step-by-step solution

Step 1: Recall the defining equation for mutual inductance

The mutual inductance $M$ between two coils is defined as:

where:

• $N_2 \Delta \Phi_{21}$ = flux linkage in coil 2 due to current in coil 1
• $\Delta I_1$ = change in current in coil 1

Step 2: Identify the given values

• Mutual inductance, $M = 1.5\ \text{H}$
• Change in current in the first coil, $\Delta I_1 = 20\ \text{A} - 0\ \text{A} = 20\ \text{A}$
• Time taken, $t = 0.5\ \text{s}$ (not needed for flux linkage — only needed if we wanted induced emf)

Step 3: Rearrange the formula to find flux linkage

From the definition:

Step 4: Substitute and calculate …

Here are the most common mistakes students make on this Mutual Inductance problem, and how to avoid each one.

Mistake 1: Confusing Flux Linkage with Induced EMF

• The Mistake: Students often calculate the induced EMF ($\mathcal{E} = -M \frac{dI}{dt}$) and then stop, thinking that is the answer. The question asks for the change in flux linkage ($\Delta \Phi$), not the voltage.
• Why it happens: The formula for induced EMF is very prominent, and the negative sign makes it look like the "main" result. Students forget that flux linkage is the cause of the EMF, not the EMF itself.
• How to Avoid:
  • Read the question twice. Underline the exact quantity asked: "change of flux linkage."
  • Recall the definition: Mutual inductance $M$ is defined as the flux linkage in one coil per unit current in the other.

Therefore, the **change** in flux linkage is directly:

- **Don't touch the EMF formula** unless the question specifically asks for induced voltage.

Mistake 2: Forgetting the "Change" ($\Delta$)

• The Mistake: Students plug in the final current value (20 A) directly into the formula, ignoring the initial current (0 A). They calculate $1.5 \times 20 = 30$, missing the fact that it's a change.
• Why it happens: The problem gives a clear "from 0 to 20 A," but in the rush to solve, students treat the final value as the only value.
• How to Avoid:
  • Always write the formula with $\Delta$ explicitly:

- **Calculate $\Delta I$ separately:**

- **Check your work:** If the current started at 10 A and went to 20 A, would your answer change? If your method doesn't account for that, you're doing it wrong.

Mistake 3: Unnecessarily Involving Time ($t$)

• The Mistake: Students see "0.5 s" and immediately calculate the rate of change of current ($dI/dt = 20 / 0.5 = 40 \ \text{A/s}$). They then multiply by $M$ to get an EMF, and then try to convert that back to flux linkage, often making an error. …