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`.
Emf induced in the smaller loop when x starts increasing at a rate v.

Options

A. `(3mu_(0)IvR^(2)r^(2))/(4pix^(4))`
B. `(6mu_(0)IvR^(2)r^(2))/(x^(4))`
C. `(3mu_(0)vR^(2)r^(2))/(2x^(4))`
D. `(3mu_(0)IvR^(2)r^(2))/(2pix^(4))`

Explanation

To solve the problem of finding the induced EMF in the smaller loop when the distance \( x \) starts increasing at a rate \( v \), we can follow these steps:

### Step 1: Determine the Magnetic Field (B) at the Location of the Smaller Loop
The magnetic field \( B \) at a distance \( x \) from a circular loop of radius \( R \) carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]
Since \( x \gg R \), we can simplify this to:
\[ B \approx \frac{\mu_0 I R}{2 x^3} \]

### Step 2: Calculate the Magnetic Flux (\( \Phi \)) through the Smaller Loop
The magnetic flux \( \Phi \) through the smaller loop of radius \( r \) is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the smaller loop:
\[ A = \pi r^2 \]
Substituting the expression for \( B \):
\[ \Phi = \left(\frac{\mu_0 I R}{2 x^3}\right) \cdot (\pi r^2) = \frac{\mu_0 I R \pi r^2}{2 x^3} \]

### Step 3: Differentiate the Flux to Find the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF (\( \mathcal{E} \)) is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Now we differentiate \( \Phi \) with respect to time \( t \):
\[ \mathcal{E} = -\frac{d}{dt}\left(\frac{\mu_0 I R \pi r^2}{2 x^3}\right) \]
Since \( \mu_0, I, R, \pi, \) and \( r \) are constants, we can focus on differentiating \( \frac{1}{x^3} \):
\[ \frac{d}{dt}\left(\frac{1}{x^3}\right) = -\frac{3}{x^4} \frac{dx}{dt} \]
Thus, we have:
\[ \mathcal{E} = -\left(\frac{\mu_0 I R \pi r^2}{2}\right) \left(-\frac{3}{x^4} \frac{dx}{dt}\right) = \frac{3\mu_0 I R \pi r^2}{2 x^4} \frac{dx}{dt} \]

### Step 4: Substitute the Rate of Change of Distance
Given that \( \frac{dx}{dt} = v \), we substitute this into our expression for EMF:
\[ \mathcal{E} = \frac{3\mu_0 I R \pi r^2}{2 x^4} v \]

### Final Answer
Thus, the induced EMF in the smaller loop is:
\[ \mathcal{E} = \frac{3\mu_0 I R \pi r^2 v}{2 x^4} \]