There is one conducting circular loop of radius R, which carries current I. There is another conducting circular loop of radius r placed coaxially at a distance x from it. Here `r lt lt R` and `x gt gt R`.
Magnetic flux linked smaller coil due to current in bigger coil is

Options

A. `(mu_(0)IR^(2)pir^(2))/(2x^(3))`
B. `(mu_(0)IR^(2)pir^(2))/(x^(3))`
C. `(2mu_(0)IR^(2)pir^(2))/(x^(3))`
D. `(mu_(0)IR^(2)r^(2))/(4pix^(3))`

Explanation

To solve the problem, we need to find the magnetic flux linked with the smaller coil due to the current in the larger coil. Here’s a step-by-step solution:

### Step 1: Find the Magnetic Field due to the Larger Coil
The magnetic field \( B \) at a distance \( x \) from a circular loop of radius \( R \) carrying a current \( I \) can be calculated using the formula:

\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

Given that \( x \gg R \), we can simplify this formula. Under this condition, \( R^2 + x^2 \approx x^2 \), so:

\[ B \approx \frac{\mu_0 I R^2}{2x^3} \]

### Step 2: Calculate the Area of the Smaller Coil
The area \( A \) of the smaller coil (which has a radius \( r \)) is given by:

\[ A = \pi r^2 \]

### Step 3: Calculate the Magnetic Flux Linked with the Smaller Coil
The magnetic flux \( \Phi \) linked with the smaller coil can be calculated using the formula:

\[ \Phi = B \cdot A \]

Substituting the expressions we derived for \( B \) and \( A \):

\[ \Phi = \left(\frac{\mu_0 I R^2}{2x^3}\right) \cdot (\pi r^2) \]

### Step 4: Final Expression for the Magnetic Flux
Thus, the magnetic flux linked with the smaller coil due to the current in the larger coil is:

\[ \Phi = \frac{\mu_0 \pi I R^2 r^2}{2x^3} \]

### Summary
The magnetic flux linked with the smaller coil is given by:

\[ \Phi = \frac{\mu_0 \pi I R^2 r^2}{2x^3} \]

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