A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn `(b gtgt 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 :

Options

A. `(mu_(0))/(4pi) 8sqrt(2) (b^(2))/(a)`
B. `(mu_(0))/(4pi)(8sqrt(2))/(a)`
C. `(mu_(0))/(4pi)(8sqrt(2))/(b)`
D. `(mu_(0))/(4pi)8sqrt(2)(a^(2))/(b)`

Answer

To find the coefficient of mutual inductance \( M \) between a small square loop of side \( a \) and a larger square loop of side \( b \) (where \( b \gg a \)), we can follow these steps:

### Step 1: Determine the Magnetic Field at the Center of the Larger Loop
The magnetic field \( B \) at the center of a square loop carrying current \( I \) can be calculated using the formula for the magnetic field due to a finite wire. For a square loop, we consider each side contributing to the magnetic field at the center.

The magnetic field due to one side of the square loop is given by:
\[ B_1 = \frac{\mu_0 I}{4\pi} \left( \frac{\sin \theta_1 + \sin \theta_2}{r} \right) \]
where \( r \) is the perpendicular distance from the wire to the center of the loop, and \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the ends of the wire at the center.

For our case, since the distance from the center to the wire is \( \frac{b}{2} \) and the angles are both \( 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus, the magnetic field due to one side becomes:
\[ B_1 = \frac{\mu_0 I}{4\pi} \left( \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{b}{2}} \right) = \frac{\mu_0 I}{4\pi} \left( \frac{2/\sqrt{2}}{b/2} \right) = \frac{\mu_0 I \sqrt{2}}{2\pi b} \]

Since there are four sides contributing to the total magnetic field:
\[ B = 4B_1 = 4 \left( \frac{\mu_0 I \sqrt{2}}{2\pi b} \right) = \frac{2\sqrt{2} \mu_0 I}{\pi b} \]

### Step 2: Calculate the Magnetic Flux Through the Smaller Loop
The magnetic flux \( \Phi \) through the smaller loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the smaller loop. The area \( A \) of the smaller square loop is \( a^2 \):
\[ \Phi = \left( \frac{2\sqrt{2} \mu_0 I}{\pi b} \right) \cdot a^2 = \frac{2\sqrt{2} \mu_0 I a^2}{\pi b} \]

### Step 3: Determine the Mutual Inductance
The mutual inductance \( M \) is defined as:
\[ M = \frac{\Phi}{I} \]
Substituting the expression for \( \Phi \):
\[ M = \frac{\frac{2\sqrt{2} \mu_0 I a^2}{\pi b}}{I} = \frac{2\sqrt{2} \mu_0 a^2}{\pi b} \]

### Final Answer
Thus, the coefficient of mutual inductance between the two loops is:
\[ \boxed{M = \frac{2\sqrt{2} \mu_0 a^2}{\pi b}} \]