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 current in the bigger loop is changed at a rate of `alpha` is.

Options

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

Explanation

To find the induced EMF in the smaller loop when the current in the larger loop is changed, we will follow these steps:

### Step 1: Understand the Problem
We have two circular loops: a larger loop with radius \( R \) carrying a current \( I \), and a smaller loop with radius \( r \) placed at a distance \( x \) from the larger loop. We know that \( r \ll R \) and \( x \gg R \). The current in the larger loop is changing at a rate of \( \alpha \) (i.e., \( \frac{dI}{dt} = \alpha \)).

### Step 2: Formula for Induced EMF
The induced EMF (\( \mathcal{E} \)) in the smaller loop can be calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux through the smaller loop.

### Step 3: Calculate Magnetic Flux
The magnetic flux \( \Phi \) through the smaller loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the smaller loop and \( B \) is the magnetic field at the location of the smaller loop due to the larger loop.

The area \( A \) of the smaller loop is:
\[ A = \pi r^2 \]

### Step 4: Find the Magnetic Field \( B \)
The magnetic field \( B \) at a distance \( x \) from the center of the larger loop can be approximated (since \( x \gg R \)) as:
\[ B = \frac{\mu_0 I R}{2 (R^2 + x^2)^{3/2}} \approx \frac{\mu_0 I R}{2 x^3} \]
This approximation is valid because \( R \) is much smaller than \( x \).

### Step 5: Substitute \( B \) into the Flux Equation
Substituting \( B \) into the flux equation, we get:
\[ \Phi = B \cdot A = \left(\frac{\mu_0 I R}{2 x^3}\right) \cdot (\pi r^2) = \frac{\mu_0 \pi r^2 I R}{2 x^3} \]

### Step 6: Differentiate Flux with Respect to Time
Now, we differentiate \( \Phi \) with respect to time to find \( \frac{d\Phi}{dt} \):
\[ \frac{d\Phi}{dt} = \frac{\mu_0 \pi r^2 R}{2 x^3} \cdot \frac{dI}{dt} \]
Since \( \frac{dI}{dt} = \alpha \), we have:
\[ \frac{d\Phi}{dt} = \frac{\mu_0 \pi r^2 R}{2 x^3} \cdot \alpha \]

### Step 7: Calculate the Induced EMF
Finally, substituting \( \frac{d\Phi}{dt} \) into the EMF equation:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\left(\frac{\mu_0 \pi r^2 R}{2 x^3} \cdot \alpha\right) \]
Taking the magnitude, we find:
\[ \mathcal{E} = \frac{\mu_0 \pi r^2 R \alpha}{2 x^3} \]

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

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