© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Physical Setup
There are two coplanar square loops, a larger one of side length $L$ (with $L \gg l$) and a smaller one of side length $l$. They share the same center. We want to find the mutual inductance $M$ of this system, which relates the current in the larger loop to the magnetic flux threading the smaller loop.
Step 2: Recall the Definition of Mutual Inductance
Mutual inductance $M$ between two circuits is defined by the relation
$ \phi = M \, i,
$
where $i$ is the current in one circuit (here, the larger loop), and $\phi$ is the resulting magnetic flux through the other circuit (the smaller loop). Our goal is to compute $M$.
Step 3: Approximate the Magnetic Field at the Center
Because $L \gg l$, each side of the larger square can be approximated as a straight current-carrying wire whose midpoint is near the center of the small loop. The magnetic field due to one long straight wire of current $i$ at a perpendicular distance $R$ is given by
$ B = \frac{\mu_{0}\,i}{2\pi R}.
$
However, in this geometry, each side of length $L$ contributes to the field at the center under some angle. Furthermore, the distance from the center of the big square to each of its sides is approximately $\frac{L}{2}$.
Step 4: Account for All Four Sides of the Larger Loop
Each side contributes a magnetic field component at the center. From symmetry, the vertical components of opposite sides cancel, and the horizontal components from each side add up, taking into account the angles. A factor of $\sin 45^\circ$ often enters if we consider the field at corners of the small loop. When fully summed, for all four sides, the net field at the center can be shown to produce a factor of
$ \left[\sin 45^\circ + \sin 45^\circ\right] = 2 \sin 45^\circ = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}.
$
Putting these details together, the effective magnetic field at the center (to be multiplied by the area of the small loop) often emerges as
$ B_{\text{eff}} \propto \frac{2\sqrt{2}\,\mu_{0}\,i}{\pi\,L}.
$
Step 5: Calculate the Magnetic Flux Through the Small Loop
The magnetic flux $\phi$ through the small square loop is given by
$ \phi = B_{\text{eff}} \times (\text{Area of the small loop}).
$
Since the area of the small loop is $l^{2}$, we write
$ \phi = \left(\frac{2\sqrt{2}\,\mu_{0}\,i}{\pi\,L}\right) \times l^{2}.
$
Step 6: Relate Flux to Mutual Inductance
By definition, $ \phi = M\,i $. Therefore,
$ M = \frac{\phi}{i}
= \frac{\frac{2\sqrt{2}\,\mu_{0}\,i}{\pi\,L} \times l^{2}}{i}
= \frac{2\sqrt{2}\,\mu_{0}\,l^{2}}{\pi\,L}.
$
Step 7: State the Final Answer
Thus, the mutual inductance of the system is
$ \displaystyle M = \frac{2 \sqrt{2}\,\mu_{0}\,l^{2}}{\pi\,L}.
$
