Two parallel wires carrying current 1 A and 3 A respectively are 1 m apart. If the current is in opposite direction, the force per unit length on both the wires will be:

Options

A. `6 xx 10^(-5)` repulsive
B. `6 xx 10^(-5)(N)/(m)` attractive
C. `6 xx 10^(-7)(N)/(m)` repulsive
D. `6 xx 10^(-5)(N)/(m)` attractive

Explanation

To find the force per unit length on two parallel wires carrying currents in opposite directions, we can use the formula for the magnetic force between two parallel currents. The formula for the force per unit length (F/L) between two parallel wires is given by:

\[ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \]

where:
- \( F/L \) is the force per unit length,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I_1 \) and \( I_2 \) are the currents in the wires,
- \( d \) is the distance between the wires.

### Step 1: Identify the values
- \( I_1 = 1 \, \text{A} \)
- \( I_2 = 3 \, \text{A} \)
- \( d = 1 \, \text{m} \)

### Step 2: Substitute the values into the formula
Substituting the values into the formula, we have:

\[ F/L = \frac{(4\pi \times 10^{-7}) \times (1) \times (3)}{2\pi \times 1} \]

### Step 3: Simplify the equation
Now, simplify the equation:

\[ F/L = \frac{4\pi \times 10^{-7} \times 3}{2\pi} \]

The \( \pi \) cancels out:

\[ F/L = \frac{4 \times 10^{-7} \times 3}{2} \]

### Step 4: Calculate the result
Calculating the above expression:

\[ F/L = \frac{12 \times 10^{-7}}{2} = 6 \times 10^{-7} \, \text{N/m} \]

### Step 5: Determine the direction of the force
Since the currents are in opposite directions, the force will be repulsive. Thus, the force per unit length on both wires is \( 6 \times 10^{-7} \, \text{N/m} \).

### Final Answer
The force per unit length on both wires is \( 6 \times 10^{-7} \, \text{N/m} \).

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