Two parallel, long wires carry currents `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 10 `mu`T. If the direction of `i_(2)` is reversed, the field becomes 30 `mu`T. Find the ratio `i_(1)/i_(2)`.

Answer

To solve the problem, we need to analyze the magnetic fields produced by the two parallel wires carrying currents \(i_1\) and \(i_2\). We will use the formula for the magnetic field due to a long straight current-carrying wire.

### 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: Analyze the Situation When Currents are in the Same Direction
When both currents \(i_1\) and \(i_2\) flow in the same direction, the magnetic fields at the midpoint (let's call it point O) will oppose each other. The total magnetic field \(B_{same}\) at point O can be expressed as:
\[ B_{same} = B_{i_1} - B_{i_2} \]
Given that the magnetic field at this point is \(10 \, \mu T\), we can write:
\[ B_{i_1} - B_{i_2} = 10 \, \mu T \]

### Step 3: Calculate the Magnetic Fields at Midpoint
Since the distance from each wire to point O is \(R\), we can substitute the magnetic field expressions:
\[ \frac{\mu_0 i_1}{2\pi R} - \frac{\mu_0 i_2}{2\pi R} = 10 \, \mu T \]
Factoring out \(\frac{\mu_0}{2\pi R}\):
\[ \frac{\mu_0}{2\pi R} (i_1 - i_2) = 10 \, \mu T \]

### Step 4: Analyze the Situation When the Direction of \(i_2\) is Reversed
When the direction of \(i_2\) is reversed, the magnetic fields at point O will add up:
\[ B_{reverse} = B_{i_1} + B_{i_2} \]
Given that the magnetic field at this point is \(30 \, \mu T\), we can write:
\[ B_{i_1} + B_{i_2} = 30 \, \mu T \]
Substituting the magnetic field expressions again:
\[ \frac{\mu_0 i_1}{2\pi R} + \frac{\mu_0 i_2}{2\pi R} = 30 \, \mu T \]
Factoring out \(\frac{\mu_0}{2\pi R}\):
\[ \frac{\mu_0}{2\pi R} (i_1 + i_2) = 30 \, \mu T \]

### Step 5: Set Up the Equations
Now we have two equations:
1. \(\frac{\mu_0}{2\pi R} (i_1 - i_2) = 10\)
2. \(\frac{\mu_0}{2\pi R} (i_1 + i_2) = 30\)

Let \(k = \frac{\mu_0}{2\pi R}\). Then we can rewrite the equations as:
1. \(k(i_1 - i_2) = 10\)
2. \(k(i_1 + i_2) = 30\)

### Step 6: Solve the Equations
From the first equation:
\[ i_1 - i_2 = \frac{10}{k} \]
From the second equation:
\[ i_1 + i_2 = \frac{30}{k} \]

Now, we can add these two equations:
\[ (i_1 - i_2) + (i_1 + i_2) = \frac{10}{k} + \frac{30}{k} \]
This simplifies to:
\[ 2i_1 = \frac{40}{k} \implies i_1 = \frac{20}{k} \]

Now, subtract the first equation from the second:
\[ (i_1 + i_2) - (i_1 - i_2) = \frac{30}{k} - \frac{10}{k} \]
This simplifies to:
\[ 2i_2 = \frac{20}{k} \implies i_2 = \frac{10}{k} \]

### Step 7: Find the Ratio
Now we can find the ratio \( \frac{i_1}{i_2} \):
\[ \frac{i_1}{i_2} = \frac{\frac{20}{k}}{\frac{10}{k}} = \frac{20}{10} = 2 \]

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

---