Animated Solution for Physics - Electromagnetic Induction: A circular wire loop of radius $R$ is placed in the $x$-$y$ plane centred at the origin $O$. A square loop of side $a$ ($a\ll R$) having two turns is placed with its centre at $z=\sqrt{3}R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of $45^{\circ }$ with respect to the Z-axis. If the mutual inductance between the loops is given by $\frac{{\mu }_0{a}^2}{2^{p/2}R}$, then the value of $p$ is

Imagine you have two completely separate circuits. No wires connect them, no physical contact exists. Yet, when you change the current in one, a mysterious "ghost" current suddenly appears in the other! This isn't magic; it's the beautiful phenomenon of mutual inductance.

In this problem, we are going to explore exactly how two such loops interact. We have a large circular loop sitting peacefully in the $xy$-plane, and a tiny square loop hovering above it on the $z$-axis. Our mission? To find out exactly how much "influence" the large loop has on the small one.

Let's break down the geometry of our setup.

We have a circular wire loop of radius $R$ centered at the origin $O$ in the $xy$-plane.

Hovering above it, at a height of $z=\sqrt{3}R$ along the $z$-axis, is a small square loop of side $a$.

The problem gives us a crucial piece of information: $a\ll R$. This means the square loop is so tiny compared to the circular loop that we can assume the magnetic field produced by the circular loop is uniform over the entire area of the square loop. This is a classic physics approximation that saves us from a nightmare of complex integration!

Furthermore, the square loop isn't just sitting flat. It's tilted! The plane of the square loop makes an angle of $45^{\circ }$ with the $z$-axis. This means its area vector (the normal to its surface) also makes an angle of $45^{\circ }$ with the $z$-axis.

Magnetic Field on the Axis

To find the mutual inductance, we need to know how much magnetic flux the first loop sends through the second loop.

Let's imagine a steady current $I$ flowing through the large circular loop. This current creates a magnetic field that permeates the space around it.

We need to find the strength of this magnetic field exactly at the location of the square loop, which is on the $z$-axis.

Do you remember the formula for the magnetic field on the axis of a circular current-carrying loop? It's one of those fundamental equations you must have at your fingertips:

This magnetic field points straight up along the $z$-axis.

Now, let's plug in the specific location of our square loop. We know it's located at $z=\sqrt{3}R$.

Let's substitute this into our magnetic field equation:

Let's simplify the denominator. The term $(\sqrt{3}R{)}^2$ simply becomes $3R^2$.

Adding this to the $R^2$ inside the parenthesis gives us $4R^2$.

So, our equation becomes:

Now, what is $(4R^2{)}^{3/2}$? We can think of this as taking the square root first, and then cubing the result. The square root of $4R^2$ is $2R$. Cubing $2R$ gives us $8R^3$.

Substituting this back into our equation:

This is the uniform magnetic field passing through the location of our tiny square loop!

Now that we have the magnetic field, we need to calculate the magnetic flux $\varphi$ passing through the square loop.

The formula for magnetic flux is:

Let's identify each of these terms for our specific square loop: 1. $N$ (Number of turns): The problem states the square loop has two turns, so $N=2$. 2. $B$ (Magnetic Field): We just calculated this! $B=\frac{{\mu }_{0}I}{16R}$. 3. $A$ (Area): The loop is a square of side $a$, so its area is $A=a^2$. 4. $\theta$ (Angle): This is the angle between the magnetic field vector and the area vector of the loop. Since the magnetic field points along the $z$-axis, and the plane of the loop makes a $45^{\circ }$ angle with the $z$-axis, the area vector also makes a $45^{\circ }$ angle with the magnetic field. So, $\theta =45^{\circ }$.

Let's substitute all these values into our flux equation:

We know that $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$.

Let's simplify this expression. The $2$ in the numerator and the $16$ in the denominator cancel out to leave an $8$ in the denominator.

To match the format given in the question, we need to express the denominator as a power of $2$.

We know that $8=2^3$ and $\sqrt{2}=2^{1/2}$.

Multiplying them together: $8\sqrt{2}=2^3\cdot 2^{1/2}=2^{3+1/2}=2^{7/2}$.

So, our final expression for the magnetic flux is:

Finding 'p'

We are almost there! The mutual inductance $M$ is defined as the total magnetic flux through the second loop divided by the current in the first loop:

Substituting our expression for flux:

The current $I$ cancels out beautifully, leaving us with:

The problem states that the mutual inductance is given by the expression:

By simply comparing our calculated expression with the given expression, we can clearly see that:

And there we have it! By breaking down the problem into logical steps—finding the magnetic field, calculating the flux, and applying the definition of mutual inductance—we've arrived at the correct answer.

This problem is a fantastic example of how different concepts in electromagnetism come together. Keep practicing, and soon these derivations will feel like second nature!