Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P, a magn

Q: Why is the mutual inductance the same in both cases?

This is due to the Reciprocity Theorem. Mutual inductance depends purely on the geometric arrangement (size, shape, number of turns, and relative position) of the two coils, not on which coil is carrying the current.

Q: What if the coils had a different number of turns?

The formula is $N_2 \Phi_2 = M I_1$. If the number of turns $N_1$ and $N_2$ were different and not equal to 1, you would need to account for them. However, the problem implies we are dealing with total flux linkages or single-turn coils, so $M = \frac{\Phi_Q}{I_P} = \frac{\Phi_P}{I_Q}$ holds.

Q: Does the distance between the coils affect the answer?

Yes, the distance affects the value of $M$. However, since the coils are 'separated by some distance' that remains constant throughout the problem, $M$ remains constant. We don't need to know the exact distance, just that it doesn't change.

Q: Why did we convert $0.667 \times 10^{-3}$ to $6.67 \times 10^{-4}$?

It's standard scientific notation to have one non-zero digit before the decimal point. Also, looking at the options provided in the question, they are formatted this way, so we match our answer to the options.

Q: What would happen if we placed a magnetic material between the coils?

A magnetic material (like an iron core) would increase the magnetic coupling between the coils, significantly increasing the mutual inductance $M$. The new $M$ would still obey the reciprocity theorem, but its value would be higher.

Question

Two coils $P$ and $Q$ are separated by some distance. When a current of $3	ext{ A}$ flows through coil $P$, a magnetic flux of $10^{-3}	ext{ Wb}$ passes through $Q$. No current is passed through $Q$. When no current passes through $P$ and a current of $2	ext{ A}$ passes through $Q$, the flux through $P$ is

Options:

(A) $6.67 	imes 10^{-3}	ext{ Wb}$

(B) $6.67 	imes 10^{-4}	ext{ Wb}$

(C) $3.67 	imes 10^{-3}	ext{ Wb}$

(D) $3.67 	imes 10^{-4}	ext{ Wb}$

Sigma Percentile · Accepted Answer

The concept of mutual inductance is one of the most elegant phenomena in electromagnetism. It describes how a changing current in one coil can induce an electromotive force (EMF), and consequently a magnetic flux, in a completely separate, nearby coil. But what happens when we reverse the roles of the coils? This problem beautifully illustrates the principle of reciprocity.

### Analyzing the Setup

Imagine two coils, $P$ and $Q$, placed at a fixed distance from each other. They are magnetically coupled, meaning the magnetic field lines generated by one coil pass through the area of the other.

In our first scenario, a current $I_P = 3	ext{ A}$ flows through coil $P$. This current generates a magnetic field, and a portion of this field links with coil $Q$. The problem states that the magnetic flux passing through coil $Q$ is $\Phi_Q = 10^{-3}	ext{ Wb}$.

The relationship between the flux in one coil and the current in the other is governed by the mutual inductance, $M$. Mathematically, this is expressed as:
$$ \Phi_Q = M_{PQ} \cdot I_P $$
where $M_{PQ}$ is the mutual inductance of coil $Q$ with respect to coil $P$.

### The Master Equation: Reciprocity Theorem

Before we calculate anything, we must invoke a powerful rule: the **Reciprocity Theorem of Mutual Inductance**. This theorem states that the mutual inductance between two coils is a shared property that depends only on their geometric arrangement (their shapes, sizes, number of turns, and relative orientation). It does not depend on which coil is acting as the source and which is the receiver.

Therefore, the mutual inductance of $Q$ with respect to $P$ is exactly equal to the mutual inductance of $P$ with respect to $Q$:
$$ M_{PQ} = M_{QP} = M $$

This is the crucial insight needed to solve the problem!

### Calculating the Mutual Inductance

Let's use the data from the first scenario to find this universal constant $M$ for our system. Substituting the known values into our flux equation:
$$ 10^{-3} = M \cdot 3 $$

Solving for $M$, we get:
$$ M = \frac{10^{-3}}{3}	ext{ H} $$

We now have the mutual inductance of the system. Because the coils haven't moved, this value remains constant for any current we pass through either coil.

### Final Calculation: Reversing the Roles

Now, we flip the script. We turn off the current in coil $P$ ($I_P = 0$) and instead pass a current $I_Q = 2	ext{ A}$ through coil $Q$. We need to find the new magnetic flux, $\Phi_P$, passing through coil $P$.

Using the same principle of mutual inductance, the equation for this new scenario is:
$$ \Phi_P = M_{QP} \cdot I_Q $$

Since we know $M_{QP} = M$, we can substitute the value we just calculated:
$$ \Phi_P = \left(\frac{10^{-3}}{3}ight) \cdot 2 $$

$$ \Phi_P = \frac{2}{3} 	imes 10^{-3}	ext{ Wb} $$

To match the format of the given options, we convert the fraction to a decimal:
$$ \Phi_P \approx 0.667 	imes 10^{-3}	ext{ Wb} $$

Finally, adjusting the scientific notation by moving the decimal point one place to the right:
$$ \Phi_P = 6.67 	imes 10^{-4}	ext{ Wb} $$

This perfectly matches option (b). The elegance of this problem lies in realizing that you don't need to know the complex geometry of the coils or the exact distance between them; the reciprocity theorem bridges the gap effortlessly.

Select Answer:

Visualized Solution

Visual Anchor

Logic Bridge

Raw Setup (Case 1)

Atomic Compute

Raw Setup (Case 2)

Atomic Compute

Final Answer

The Way Forward

The Sigma Insight: Inductance (Self & Mutual)

Analyzing the Setup

The Master Equation

Calculating the Mutual Inductance

Final Calculation

Similar Questions

There are two long coaxial solenoids of same length $l$ . The inner and outer coils have radii $r_{1}$ and $r_{2}$ and number of turns per unit length $n_{1}$ and $n_{2}$ , respectively. The ratio of mutual inductance to the self-inductance of the inner coil is