Three long wires are situated in a plane with equal separation. A current of 1A, 2A, 3A flows through these wires in the same direction. What is ratio of `F_(1)//F_(2)`? Where `F_(1)` & `F_(2)` are force per unit length on the wire 1 and 2 respectively:

Options

A. `7//8`
B. 1
C. `9//8`
D. None of these

Explanation

To solve the problem of finding the ratio \( \frac{F_1}{F_2} \) where \( F_1 \) and \( F_2 \) are the forces per unit length on wires 1 and 2 respectively, we can follow these steps:

### Step 1: Understand the Configuration
We have three parallel wires with currents flowing through them:
- Wire 1: Current \( I_1 = 1 \, \text{A} \)
- Wire 2: Current \( I_2 = 2 \, \text{A} \)
- Wire 3: Current \( I_3 = 3 \, \text{A} \)

The wires are equally spaced apart by a distance \( d \).

### Step 2: Calculate the Force on Wire 1
The force per unit length \( F_1 \) on wire 1 is due to the magnetic fields created by wires 2 and 3. The formula for the force per unit length between two parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( d \) is given by:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi d} \]

#### Force due to Wire 2 on Wire 1:
The force \( F_{12} \) (force on wire 1 due to wire 2) is:
\[ F_{12} = \frac{\mu_0 I_1 I_2}{2 \pi d} = \frac{\mu_0 (1)(2)}{2 \pi d} = \frac{2 \mu_0}{2 \pi d} = \frac{\mu_0}{\pi d} \]

#### Force due to Wire 3 on Wire 1:
The distance from wire 1 to wire 3 is \( 2d \). Thus, the force \( F_{13} \) (force on wire 1 due to wire 3) is:
\[ F_{13} = \frac{\mu_0 I_1 I_3}{2 \pi (2d)} = \frac{\mu_0 (1)(3)}{4 \pi d} = \frac{3 \mu_0}{4 \pi d} \]

#### Total Force on Wire 1:
Since both forces \( F_{12} \) and \( F_{13} \) act in the same direction, we can add them:
\[ F_1 = F_{12} + F_{13} = \frac{\mu_0}{\pi d} + \frac{3 \mu_0}{4 \pi d} \]
To add these fractions, we need a common denominator:
\[ F_1 = \frac{4 \mu_0}{4 \pi d} + \frac{3 \mu_0}{4 \pi d} = \frac{7 \mu_0}{4 \pi d} \]

### Step 3: Calculate the Force on Wire 2
The force \( F_2 \) on wire 2 is due to wires 1 and 3.

#### Force due to Wire 1 on Wire 2:
\[ F_{21} = \frac{\mu_0 I_2 I_1}{2 \pi d} = \frac{\mu_0 (2)(1)}{2 \pi d} = \frac{2 \mu_0}{2 \pi d} = \frac{\mu_0}{\pi d} \]

#### Force due to Wire 3 on Wire 2:
\[ F_{23} = \frac{\mu_0 I_2 I_3}{2 \pi d} = \frac{\mu_0 (2)(3)}{2 \pi d} = \frac{6 \mu_0}{2 \pi d} = \frac{3 \mu_0}{\pi d} \]

#### Total Force on Wire 2:
The forces \( F_{21} \) and \( F_{23} \) act in opposite directions, so we subtract:
\[ F_2 = F_{21} - F_{23} = \frac{\mu_0}{\pi d} - \frac{3 \mu_0}{\pi d} = -\frac{2 \mu_0}{\pi d} \]

### Step 4: Calculate the Ratio \( \frac{F_1}{F_2} \)
Now we can find the ratio:
\[ \frac{F_1}{F_2} = \frac{\frac{7 \mu_0}{4 \pi d}}{-\frac{2 \mu_0}{\pi d}} = \frac{7}{4} \cdot \left(-\frac{\pi d}{2 \mu_0}\right) = -\frac{7}{8} \]

### Final Answer
The ratio \( \frac{F_1}{F_2} \) is:
\[ \frac{F_1}{F_2} = -\frac{7}{8} \]