Two concentric coplanar circular loops of radii `r_(1)` and `r_(2)` carry currents of respectively `i_(1)` and `i_(2)` in opposite direction (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to `i_(1)` alone at the centre. if `r_(2)=2r_(1)`. the value of `i_(2)//i_(1)` is

Options

A. `2`
B. `1//2`
C. `1//4`
D. `1`

Explanation

To solve the problem, we need to find the ratio \( \frac{i_2}{i_1} \) given the conditions of the two concentric circular loops. Let's break down the solution step by step.

### Step 1: Understand the Magnetic Field due to a Single Loop
The magnetic field \( B \) at the center of a circular loop carrying current \( i \) and having radius \( r \) is given by the formula:
\[ B = \frac{\mu_0 i}{2r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Calculate the Magnetic Field due to Each Loop
For the first loop with radius \( r_1 \) and current \( i_1 \):
\[ B_1 = \frac{\mu_0 i_1}{2r_1} \]

For the second loop with radius \( r_2 = 2r_1 \) and current \( i_2 \):
\[ B_2 = \frac{\mu_0 i_2}{2r_2} = \frac{\mu_0 i_2}{2(2r_1)} = \frac{\mu_0 i_2}{4r_1} \]

### Step 3: Determine the Net Magnetic Field at the Center
Since the currents are in opposite directions, the net magnetic field \( B_{net} \) at the center of the loops is given by:
\[ B_{net} = B_1 - B_2 \]
Substituting the expressions for \( B_1 \) and \( B_2 \):
\[ B_{net} = \frac{\mu_0 i_1}{2r_1} - \frac{\mu_0 i_2}{4r_1} \]

### Step 4: Set Up the Equation Based on Given Condition
According to the problem, the net magnetic field at the center is half of the magnetic field due to \( i_1 \) alone:
\[ B_{net} = \frac{1}{2} B_1 = \frac{1}{2} \left( \frac{\mu_0 i_1}{2r_1} \right) = \frac{\mu_0 i_1}{4r_1} \]

### Step 5: Equate the Two Expressions for \( B_{net} \)
Now we can equate the two expressions for \( B_{net} \):
\[ \frac{\mu_0 i_1}{2r_1} - \frac{\mu_0 i_2}{4r_1} = \frac{\mu_0 i_1}{4r_1} \]

### Step 6: Simplify the Equation
Dividing through by \( \mu_0 \) and multiplying by \( 4r_1 \) to eliminate the fractions:
\[ 2i_1 - i_2 = i_1 \]
This simplifies to:
\[ 2i_1 - i_1 = i_2 \quad \Rightarrow \quad i_2 = i_1 \]

### Step 7: Find the Ratio \( \frac{i_2}{i_1} \)
Thus, the ratio of the currents is:
\[ \frac{i_2}{i_1} = \frac{i_1}{i_1} = 1 \]

### Final Answer
The value of \( \frac{i_2}{i_1} \) is \( 1 \).

---