There is a conducting ring of radius `R`. Another ring having current `i` and radius `r (r lt lt R)` is kept on the axis of bigger ring such that its center lies on the axis of bigger ring at a distance `x` from the center of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring is

Options

A. (a)`(mu_(0)piR^(2)r^(2))/((R^(2)+x^(2))^(3//2)`
B. (b)`(mu_(0)piR^(2)r^(2))/((R^(2)+x^(4))^(3//2)`
C. (c)`(mu_(0)piR^(2)r^(2))/(16(R^(2)+x^(2))^(3//2)`
D. (d)`(mu_(0)piR^(2)r^(2))/(2(R^(2)+x^(2))^(3//2)`

Answer

To find the mutual inductance \( M \) of the larger ring due to the smaller ring, we can follow these steps:

### Step 1: Understanding the Magnetic Field
The magnetic field \( B \) at a distance \( x \) along the axis of a circular current-carrying loop (the larger ring) is given by the formula:
\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current in the larger ring,
- \( R \) is the radius of the larger ring,
- \( x \) is the distance from the center of the larger ring to the smaller ring.

### Step 2: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the smaller ring of radius \( r \) placed at a distance \( x \) from the center of the larger ring can be calculated using the formula:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the smaller ring given by \( A = \pi r^2 \). Thus, we have:
\[ \Phi = B \cdot \pi r^2 \]

### Step 3: Substitute the Magnetic Field into the Flux Equation
Substituting the expression for \( B \) into the flux equation gives:
\[ \Phi = \left(\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\right) \cdot \pi r^2 \]
This simplifies to:
\[ \Phi = \frac{\mu_0 I R^2 \pi r^2}{2(R^2 + x^2)^{3/2}} \]

### Step 4: Relate Flux to Mutual Inductance
The mutual inductance \( M \) is defined by the relationship:
\[ \Phi = M I \]
where \( I \) is the current in the smaller ring. Rearranging gives:
\[ M = \frac{\Phi}{I} \]

### Step 5: Substitute for \( \Phi \) and Simplify
Substituting the expression for \( \Phi \) into the equation for \( M \):
\[ M = \frac{\frac{\mu_0 I R^2 \pi r^2}{2(R^2 + x^2)^{3/2}}}{I} \]
The \( I \) cancels out, leading to:
\[ M = \frac{\mu_0 R^2 \pi r^2}{2(R^2 + x^2)^{3/2}} \]

### Step 6: Final Expression for Mutual Inductance
Thus, the final expression for the mutual inductance \( M \) of the larger ring due to the smaller ring is:
\[ M = \frac{\mu_0 \pi r^2 R^2}{2(R^2 + x^2)^{3/2}} \]