A large circular loop of radius 1 metre has total resistance `4Omega`. A variable current is folwing through the loop. The current at any instant is given by equation l = lA, where `i` is time in second. A very small circular loop or radius 1 cm having resistance 3.14 `Omega` is placed coaxially with the larger loop at a distance of `sqrt3` metre from the centre of larger loop. If induced current in smaller loop is `(mu_(0)x xx 10^(-4))/(16)` ampere then, find x.

Answer

To solve the problem, we need to find the value of \( x \) given the conditions of the large and small circular loops. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- We have a large circular loop (L-loop) with:
- Radius \( R_L = 1 \, \text{m} \)
- Resistance \( R_L = 4 \, \Omega \)
- Current \( I(t) = I_a \) (where \( I_a \) is the current at time \( t \)).

- A small circular loop (S-loop) with:
- Radius \( R_S = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Resistance \( R_S = 3.14 \, \Omega \)

- The distance from the center of the L-loop to the center of the S-loop is \( d = \sqrt{3} \, \text{m} \).

### Step 2: Calculate the Magnetic Field at the Center of the S-loop
The magnetic field \( B_S \) at a distance \( d \) from the center of a circular loop carrying current \( I \) is given by the formula:

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

For our case:
- \( R = R_L = 1 \, \text{m} \)
- \( d = \sqrt{3} \, \text{m} \)

Substituting the values into the formula:

\[ B_S = \frac{\mu_0 I_a (1^2)}{(1^2 + (\sqrt{3})^2)^{3/2}} = \frac{\mu_0 I_a}{(1 + 3)^{3/2}} = \frac{\mu_0 I_a}{4^{3/2}} = \frac{\mu_0 I_a}{8} \]

### Step 3: Calculate the Induced EMF in the S-loop
The induced EMF \( E \) in the S-loop is given by Faraday's law of electromagnetic induction:

\[ E = -\frac{d\Phi}{dt} \]

Where \( \Phi \) is the magnetic flux through the S-loop. The magnetic flux \( \Phi \) is given by:

\[ \Phi = B_S \cdot A_S \]

Where \( A_S \) is the area of the S-loop:

\[ A_S = \pi R_S^2 = \pi (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \]

Thus, the magnetic flux becomes:

\[ \Phi = \frac{\mu_0 I_a}{8} \cdot \pi \times 10^{-4} \]

Now, differentiate \( \Phi \) with respect to time \( t \):

\[ E = \frac{d\Phi}{dt} = \frac{\pi \mu_0}{8} \frac{dI_a}{dt} \]

Since \( I_a = I_a \) (a constant factor), we can express \( \frac{dI_a}{dt} = 1 \) (assuming \( I_a \) changes linearly with time).

### Step 4: Calculate the Induced Current in the S-loop
The induced current \( I \) in the S-loop is given by:

\[ I = \frac{E}{R_S} \]

Substituting \( E \):

\[ I = \frac{\frac{\pi \mu_0}{8}}{3.14} = \frac{\mu_0}{8} \]

### Step 5: Compare with Given Induced Current Expression
According to the problem, the induced current is given as:

\[ I = \frac{\mu_0 x \times 10^{-4}}{16} \]

Setting the two expressions for induced current equal:

\[ \frac{\mu_0}{8} = \frac{\mu_0 x \times 10^{-4}}{16} \]

Cancelling \( \mu_0 \) and rearranging gives:

\[ \frac{1}{8} = \frac{x \times 10^{-4}}{16} \]

Cross-multiplying yields:

\[ 16 = 8x \times 10^{-4} \]

Solving for \( x \):

\[ x = \frac{16}{8 \times 10^{-4}} = 2 \]

### Final Answer
Thus, the value of \( x \) is \( 2 \).