A small square loop of wire of side `l` is placed inside a large square loop of wire of side `L (L gt gt l)`. The loops are coplanar and their centre coincide. What is the mutual inductance of the system ?

Options

A. `2sqrt2(mu_(0))/(pi)(l^(2))/(L)`
B. `8sqrt2(mu_(0))/(pi)(l^(2))/(L)`
C. `2sqrt2(mu_(0))/(2pi)(l^(2))/(L)`
D. `2sqrt2(mu_(0)L^(2))/(pil)`

Explanation

To find the mutual inductance \( M \) of a small square loop of wire of side \( l \) placed inside a large square loop of wire of side \( L \) (where \( L \gg l \)), we can follow these steps:

### Step 1: Understanding the Setup
We have two square loops:
- A large square loop with side length \( L \).
- A small square loop with side length \( l \) placed inside the large loop, such that their centers coincide.

### Step 2: Magnetic Field at the Center of the Large Loop
When a current \( I \) flows through the large loop, it generates a magnetic field at its center. The magnetic field \( B \) due to a single side of the square loop at the center can be calculated using the formula:
\[ B = \frac{\mu_0 I}{4\pi d} \sin(\theta) \]
where \( d \) is the distance from the wire to the point where we are calculating the field, and \( \theta \) is the angle subtended at that point by the wire.

For a square loop, the angle \( \theta \) at the center from each side is \( 45^\circ \). The distance \( d \) from the center to the midpoint of a side of the large loop is \( \frac{L}{2} \).

### Step 3: Calculate the Magnetic Field Contribution
The total magnetic field at the center due to all four sides of the large loop can be calculated as:
\[ B = 4 \cdot \left( \frac{\mu_0 I}{4\pi \frac{L}{2}} \sin(45^\circ) \right) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ B = 4 \cdot \left( \frac{\mu_0 I}{4\pi \frac{L}{2}} \cdot \frac{1}{\sqrt{2}} \right) = \frac{2\sqrt{2} \mu_0 I}{\pi L} \]

### Step 4: Calculate the Magnetic Flux through the Small Loop
The magnetic flux \( \Phi \) through the small loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the small loop. The area \( A \) of the small loop is \( l^2 \), so:
\[ \Phi = \left( \frac{2\sqrt{2} \mu_0 I}{\pi L} \right) \cdot l^2 \]

### Step 5: Calculate the Mutual Inductance
The mutual inductance \( M \) is defined as:
\[ M = \frac{\Phi}{I} \]
Substituting the expression for \( \Phi \):
\[ M = \frac{\left( \frac{2\sqrt{2} \mu_0 I}{\pi L} \cdot l^2 \right)}{I} = \frac{2\sqrt{2} \mu_0 l^2}{\pi L} \]

### Final Result
Thus, the mutual inductance of the system is:
\[ M = \frac{2\sqrt{2} \mu_0 l^2}{\pi L} \]