Two long parallel wires carry currents `i_(1)` and `i_(2)` such that `i_(1) gt i_(2)`. When the currents are in the same direction, the magnetic field at a point midway between the wires is `6xx10^(-6)T`. If the direction of `i_(2)` is reversed, the field becomes `3xx10^(-5)T`. The ratio `(i_(1))/(i_(2))` is

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(3)/(2)`
D. `(1)/(5)`

Answer

To solve the problem, we will analyze the magnetic fields produced by two long parallel wires carrying currents \( i_1 \) and \( i_2 \). We will derive the ratio \( \frac{i_1}{i_2} \) based on the given magnetic field strengths in two scenarios.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field due to a Long Straight Wire:**
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

2. **Case 1: Currents in the Same Direction**
When both currents \( i_1 \) and \( i_2 \) are in the same direction, the magnetic field at the midpoint (let's call it point P) between the wires is:
\[ B = B_1 - B_2 \]
where \( B_1 \) is the magnetic field due to wire 1 and \( B_2 \) is the magnetic field due to wire 2. Therefore:
\[ B_1 = \frac{\mu_0 i_1}{2 \pi r}, \quad B_2 = \frac{\mu_0 i_2}{2 \pi r} \]
Thus, we have:
\[ B = \frac{\mu_0 i_1}{2 \pi r} - \frac{\mu_0 i_2}{2 \pi r} = \frac{\mu_0}{2 \pi r} (i_1 - i_2) \]
Given that this magnetic field equals \( 6 \times 10^{-6} \, T \), we can write:
\[ \frac{\mu_0}{2 \pi r} (i_1 - i_2) = 6 \times 10^{-6} \quad \text{(Equation 1)} \]

3. **Case 2: Current \( i_2 \) is Reversed**
When the direction of \( i_2 \) is reversed, the magnetic fields add up:
\[ B = B_1 + B_2 \]
Therefore:
\[ B = \frac{\mu_0 i_1}{2 \pi r} + \frac{\mu_0 i_2}{2 \pi r} = \frac{\mu_0}{2 \pi r} (i_1 + i_2) \]
Given that this magnetic field equals \( 3 \times 10^{-5} \, T \), we can write:
\[ \frac{\mu_0}{2 \pi r} (i_1 + i_2) = 3 \times 10^{-5} \quad \text{(Equation 2)} \]

4. **Dividing the Two Equations:**
Now, we will divide Equation 2 by Equation 1:
\[ \frac{\frac{\mu_0}{2 \pi r} (i_1 + i_2)}{\frac{\mu_0}{2 \pi r} (i_1 - i_2)} = \frac{3 \times 10^{-5}}{6 \times 10^{-6}} \]
The \( \frac{\mu_0}{2 \pi r} \) cancels out:
\[ \frac{i_1 + i_2}{i_1 - i_2} = 5 \]

5. **Cross Multiplying:**
Cross-multiplying gives us:
\[ i_1 + i_2 = 5(i_1 - i_2) \]
Expanding this:
\[ i_1 + i_2 = 5i_1 - 5i_2 \]
Rearranging terms:
\[ i_1 + i_2 + 5i_2 = 5i_1 \]
\[ 6i_2 = 4i_1 \]

6. **Finding the Ratio:**
Dividing both sides by \( i_2 \):
\[ \frac{i_1}{i_2} = \frac{6}{4} = \frac{3}{2} \]

### Final Answer:
The ratio \( \frac{i_1}{i_2} \) is \( \frac{3}{2} \).