Two parallel, long wires carry current `i_(1)` and `i_(2)` with`i_(1)gti_(2)`. When the currents are in the same direction, the magnetic field at a point midway between the wires is `30 muT`. If the direction of `i_(1)` is reversed , the field becomes `90muT`. The ratio `i_(1)//i_(2)` is

Options

A. `1//2`
B. `2`
C. `2//3`
D. `3//2`

Explanation

To solve the problem, we will analyze the magnetic fields created by two parallel wires carrying currents \(i_1\) and \(i_2\) under two different configurations.

### Step 1: Understand the Magnetic Field Due to a Current-Carrying Wire
The magnetic field \(B\) at a distance \(r\) from a long straight wire carrying current \(I\) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \(\mu_0\) is the permeability of free space.

### Step 2: Set Up the Equations for Both Cases
1. **First Case (Currents in Same Direction)**:
- The magnetic field at the midpoint between the two wires is the sum of the magnetic fields due to each wire.
- Let the distance from each wire to the midpoint be \(r\).
- The magnetic field due to wire 1 is \(B_1 = \frac{\mu_0 i_1}{2 \pi r}\) (into the page).
- The magnetic field due to wire 2 is \(B_2 = \frac{\mu_0 i_2}{2 \pi r}\) (also into the page).
- The net magnetic field \(B\) at the midpoint is:
\[ B = B_1 + B_2 = \frac{\mu_0 i_1}{2 \pi r} + \frac{\mu_0 i_2}{2 \pi r} = \frac{\mu_0}{2 \pi r} (i_1 + i_2) \]
- Given that \(B = 30 \, \mu T\), we have:
\[ \frac{\mu_0}{2 \pi r} (i_1 + i_2) = 30 \times 10^{-6} \, T \quad \text{(Equation 1)} \]

2. **Second Case (Currents in Opposite Directions)**:
- When \(i_1\) is reversed, the magnetic field due to wire 1 is now out of the page.
- The magnetic field at the midpoint is:
\[ B = B_1 - B_2 = \frac{\mu_0 i_1}{2 \pi r} - \frac{\mu_0 i_2}{2 \pi r} = \frac{\mu_0}{2 \pi r} (i_1 - i_2) \]
- Given that \(B = 90 \, \mu T\), we have:
\[ \frac{\mu_0}{2 \pi r} (i_1 - i_2) = 90 \times 10^{-6} \, T \quad \text{(Equation 2)} \]

### Step 3: Divide Equation 2 by Equation 1
To eliminate \(\frac{\mu_0}{2 \pi r}\), we can divide Equation 2 by Equation 1:
\[ \frac{i_1 - i_2}{i_1 + i_2} = \frac{90 \times 10^{-6}}{30 \times 10^{-6}} = 3 \]
This simplifies to:
\[ i_1 - i_2 = 3(i_1 + i_2) \]
Expanding this gives:
\[ i_1 - i_2 = 3i_1 + 3i_2 \]
Rearranging terms:
\[ i_1 - 3i_1 = 3i_2 + i_2 \implies -2i_1 = 4i_2 \implies \frac{i_1}{i_2} = \frac{4}{2} = 2 \]

### Final Answer
Thus, the ratio \( \frac{i_1}{i_2} \) is:
\[ \frac{i_1}{i_2} = 2:1 \]