There are two straight conductors, carrying currents `i_(1)" and " i_(2)`. Both the conductors are kept parallel to each other at a separation 2d. The magnetic field is measured at a point midway between the conductors. Magnetic field intensity at the above specified point is `20 mu`T when both the wire carry current in the same direction but it becomes `80 mu T` when the current in one of them is reversed. Assume `i_(1) gt i_(2)` and find `i_(1)//i_(2)`.

Options

A. 2
B. `5//3`
C. 4
D. `7//3`

Explanation

To solve the problem, we need to analyze the magnetic fields produced by two parallel conductors carrying currents \(i_1\) and \(i_2\) at a point midway between them. We will use the formula for the magnetic field due to a long straight conductor and apply it to both scenarios described in the question.

### Step-by-Step Solution:

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

2. **Setting Up the Problem**:
- Let the distance between the two conductors be \(2d\), so the distance from each conductor to the midpoint is \(d\).
- When both currents are in the same direction, the magnetic fields at the midpoint due to each conductor will oppose each other since they are in the same direction.

3. **Case 1: Currents in the Same Direction**:
- The magnetic field due to conductor 1 at the midpoint:
\[ B_1 = \frac{\mu_0 i_1}{2 \pi d} \]
- The magnetic field due to conductor 2 at the midpoint:
\[ B_2 = \frac{\mu_0 i_2}{2 \pi d} \]
- Since both currents are in the same direction, the net magnetic field \(B_{net1}\) at the midpoint is:
\[ B_{net1} = B_1 - B_2 = \frac{\mu_0 i_1}{2 \pi d} - \frac{\mu_0 i_2}{2 \pi d} = \frac{\mu_0}{2 \pi d} (i_1 - i_2) \]
- We know from the problem that \(B_{net1} = 20 \, \mu T\):
\[ \frac{\mu_0}{2 \pi d} (i_1 - i_2) = 20 \times 10^{-6} \]

4. **Case 2: Currents in Opposite Directions**:
- When the currents are in opposite directions, the magnetic fields will add up:
\[ B_{net2} = B_1 + B_2 = \frac{\mu_0 i_1}{2 \pi d} + \frac{\mu_0 i_2}{2 \pi d} = \frac{\mu_0}{2 \pi d} (i_1 + i_2) \]
- We know from the problem that \(B_{net2} = 80 \, \mu T\):
\[ \frac{\mu_0}{2 \pi d} (i_1 + i_2) = 80 \times 10^{-6} \]

5. **Setting Up the Equations**:
- We now have two equations:
1. \(\frac{\mu_0}{2 \pi d} (i_1 - i_2) = 20 \times 10^{-6}\) (Equation 1)
2. \(\frac{\mu_0}{2 \pi d} (i_1 + i_2) = 80 \times 10^{-6}\) (Equation 2)

6. **Solving the Equations**:
- From Equation 1:
\[ i_1 - i_2 = \frac{20 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]
- From Equation 2:
\[ i_1 + i_2 = \frac{80 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]

7. **Adding the Two Equations**:
- Adding both equations:
\[ (i_1 - i_2) + (i_1 + i_2) = \frac{20 \times 10^{-6} \cdot 2 \pi d}{\mu_0} + \frac{80 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]
\[ 2i_1 = \frac{100 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]
\[ i_1 = \frac{50 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]

8. **Subtracting the Two Equations**:
- Subtracting Equation 1 from Equation 2:
\[ (i_1 + i_2) - (i_1 - i_2) = \frac{80 \times 10^{-6} \cdot 2 \pi d}{\mu_0} - \frac{20 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]
\[ 2i_2 = \frac{60 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]
\[ i_2 = \frac{30 \times 10^{-6} \cdot 2 \pi d}{\mu_0} \]

9. **Finding the Ratio \(i_1/i_2\)**:
- Now, we can find the ratio:
\[ \frac{i_1}{i_2} = \frac{\frac{50 \times 10^{-6} \cdot 2 \pi d}{\mu_0}}{\frac{30 \times 10^{-6} \cdot 2 \pi d}{\mu_0}} = \frac{50}{30} = \frac{5}{3} \]

### Final Answer:
\[ \frac{i_1}{i_2} = \frac{5}{3} \]