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Step-by-Step Solution
Step 1: Identify the relevant physical principle
The problem involves finding the mutual inductance between two coplanar square loops (a smaller loop of side $a$ inside a larger loop of side $b \, (b \gg a)$). By definition, the mutual inductance $M$ between two circuits is given by
$$
M = \frac{\Phi}{I},
$$
where $\Phi$ is the magnetic flux through one loop due to the current $I$ in the other loop.
Step 2: Magnetic field at the center of a single square loop
When a current $I$ flows in a square loop of side $b$, the magnetic field at the center of the loop (assuming a single turn) is given by
$$
B_{\text{center}} = \frac{2\sqrt{2} \, \mu_0 \, I}{\pi \, b}.
$$
For large $b$ compared to $a$, we can approximate that the magnetic field throughout the area of the small loop is nearly the same as it is at the center of the large loop.
Step 3: Calculate the flux through the smaller loop
Since the smaller loop of side $a$ lies inside the larger loop and is coplanar, the magnetic flux $\Phi$ linked with the smaller loop is the magnetic field (from the bigger loop) times the area of the smaller loop:
$$
\Phi = B_{\text{center}} \times \text{Area of small loop}
= \left( \frac{2\sqrt{2} \, \mu_0 \, I}{\pi \, b} \right) \times a^2.
$$
Step 4: Determine the mutual inductance
By definition of mutual inductance,
$$
M = \frac{\Phi}{I}.
$$
Substituting the flux $\Phi$ calculated above and dividing by $I$ (the current in the larger loop), we get
$$
M = \frac{\left( \frac{2\sqrt{2} \, \mu_0 \, I}{\pi \, b} \right) a^2}{I}
= \frac{2\sqrt{2} \, \mu_0 \, a^2}{\pi \, b}.
$$
This can also be written equivalently as
$$
M = \frac{\mu_0}{4\pi} \, 8\sqrt{2} \, \frac{a^2}{b},
$$
matching the given correct answer.
Step 5: Final expression for the mutual inductance
Therefore, the coefficient of mutual inductance between the two loops is
$$
M = \frac{\mu_0}{4\pi}\, 8 \sqrt{2}\,\frac{a^2}{b}.
$$
Reference Diagram
