Animated Solution for Physics - Electromagnetic Induction: A small square loop of side $a$ and one turn is placed inside a larger square loop of side $b$ and one turn ($b\gg a$). The two loops are coplanar with their centres coinciding. If a current $I$ is passed in the square loop of side $b$, then the coefficient of mutual inductance between the two loops is

Have you ever wondered how two completely disconnected circuits can 'talk' to each other? Welcome to the fascinating world of mutual inductance! In this problem, we are going to explore a classic setup: a tiny square loop nestled inside a giant square loop. Let's break down the physics and the math behind this elegant phenomenon.

Two Concentric Squares

Imagine two square loops lying flat on a table, perfectly centered on each other. The outer loop is massive, with a side length of $b$, while the inner loop is tiny, with a side length of $a$. We are given a critical condition: $b\gg a$.

We send a steady current $I$ through the large outer loop. This current generates a magnetic field that permeates the space around it, including the area enclosed by the tiny inner loop. Our mission is to find the coefficient of mutual inductance, $M$, between these two loops.

By definition, mutual inductance is the ratio of the magnetic flux $\varphi$ linked with the secondary coil (the inner loop) to the current $I$ flowing through the primary coil (the outer loop):

Here is where the physics intuition kicks in. Calculating the exact magnetic flux through a square loop due to another square loop is generally a mathematical nightmare involving complex surface integrals. However, the condition $b\gg a$ is our golden ticket.

Because the inner loop is so incredibly small compared to the outer one, it effectively acts like a point at the center of the large square. The magnetic field produced by the outer loop varies wildly near its wires, but right at the center, over the tiny area $a^2$, the field is practically uniform.

Therefore, we don't need to integrate! We just need to find the magnetic field $B$ exactly at the center of the large square and multiply it by the area of the small square.

Let's find the magnetic field at the center of the large square loop. The loop consists of four identical straight wire segments. We can find the field due to one segment and then multiply by 4, thanks to the principle of superposition and the right-hand rule (which tells us all four sides produce a field pointing in the same direction at the center).

For a finite straight wire, the magnetic field at a perpendicular distance $d$ is given by:

For our square of side $b$, the distance to the center is $d=\frac{b}{2}$. The angles subtended by the ends of the wire at the center are ${\theta }_1={\theta }_2=45^{\circ }$. Substituting these values:

Now, we multiply by 4 to get the total magnetic field $B$ at the center:

To match the standard format of the options, let's multiply the numerator and denominator by 4 to keep the $\frac{{\mu }_0}{4\pi }$ term intact:

Now that we have our uniform magnetic field $B$, finding the flux $\varphi$ through the inner loop is a breeze. We just multiply the field by the area of the inner loop, which is $a^2$:

Finally, we plug this flux back into our mutual inductance formula. Notice how beautifully the current $I$ cancels out, proving that mutual inductance is purely a geometric property of the system:

And there we have it! A complex-looking problem dismantled into simple, logical steps using the power of physical approximations.