Two parallel long wires carry currents `i_(1) "and" i_(2)` with `i_(1) gt i_(2)`.When the currents are in the same direction then the magnetic field midway between the wires is `10 mu T.` when the direction of `i_(2)` is reversed ,then it becomes `40 mu T.` then ratio of `i_(1)//i_(2)` is

Options

A. `3:4`
B. `5:3`
C. `7:11`
D. `11:7`

Answer

To solve the problem, we need to analyze the magnetic fields produced by two parallel wires carrying currents \( i_1 \) and \( i_2 \).

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Between Two Wires:**
When two long parallel wires carry currents, they create magnetic fields around them. The magnetic field at a point between the wires depends on the direction of the currents.

2. **Case 1: Currents in the Same Direction**
- When both currents are in the same direction, the magnetic fields due to each wire at the midpoint will oppose each other.
- The net magnetic field \( B \) at the midpoint can be expressed as:
\[ B = B_1 - B_2 \]
- The magnetic field due to a long straight wire is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- Therefore, for the two wires, we have:
\[ B = \frac{\mu_0 i_1}{2 \pi r} - \frac{\mu_0 i_2}{2 \pi r} \]
- This simplifies to:
\[ B = \frac{\mu_0}{2 \pi r} (i_1 - i_2) \]
- Given that \( B = 10 \, \mu T = 10 \times 10^{-6} T \), we can set up our first equation:
\[ \frac{\mu_0}{2 \pi r} (i_1 - i_2) = 10 \times 10^{-6} \quad \text{(Equation 1)} \]

3. **Case 2: One Current Reversed**
- When the direction of \( i_2 \) is reversed, the magnetic fields from both wires will add up:
\[ B = B_1 + B_2 \]
- Thus, we have:
\[ B = \frac{\mu_0 i_1}{2 \pi r} + \frac{\mu_0 i_2}{2 \pi r} \]
- This simplifies to:
\[ B = \frac{\mu_0}{2 \pi r} (i_1 + i_2) \]
- Given that \( B = 40 \, \mu T = 40 \times 10^{-6} T \), we can set up our second equation:
\[ \frac{\mu_0}{2 \pi r} (i_1 + i_2) = 40 \times 10^{-6} \quad \text{(Equation 2)} \]

4. **Dividing the Two Equations:**
- Now, we can divide Equation 1 by Equation 2:
\[ \frac{\frac{\mu_0}{2 \pi r} (i_1 - i_2)}{\frac{\mu_0}{2 \pi r} (i_1 + i_2)} = \frac{10 \times 10^{-6}}{40 \times 10^{-6}} \]
- This simplifies to:
\[ \frac{i_1 - i_2}{i_1 + i_2} = \frac{1}{4} \]

5. **Cross Multiplying:**
- Cross multiplying gives:
\[ 4(i_1 - i_2) = i_1 + i_2 \]
- Expanding this leads to:
\[ 4i_1 - 4i_2 = i_1 + i_2 \]

6. **Rearranging the Equation:**
- Rearranging gives:
\[ 4i_1 - i_1 = 4i_2 + i_2 \]
- Thus:
\[ 3i_1 = 5i_2 \]

7. **Finding the Ratio:**
- Dividing both sides by \( i_2 \):
\[ \frac{i_1}{i_2} = \frac{5}{3} \]

### Final Answer:
The ratio of \( i_1 \) to \( i_2 \) is:
\[ \frac{i_1}{i_2} = \frac{5}{3} \]