Two long and parallel wires are at a distance of `0.1m` and a current of `5 A` is flowing in each of these wires. The force per unit length due to these wires will be

Options

A. `5xx10^(-5)N//m`
B. `5xx10^(-3)N//m`
C. `2.5xx10^(-5)N//m`
D. `2.5xx10^(-4)N//m`

Explanation

To find the force per unit length between two long parallel wires carrying currents, we can use the formula derived from Ampère's force law. Here’s a step-by-step solution:

### Step 1: Identify the parameters
We have two long parallel wires separated by a distance \( R = 0.1 \, \text{m} \) and carrying currents \( I_1 = 5 \, \text{A} \) and \( I_2 = 5 \, \text{A} \).

### Step 2: Write the formula for the magnetic field
The magnetic field \( B \) created by a long straight wire at a distance \( R \) from it is given by:
\[ B = \frac{\mu_0 I_1}{2 \pi R} \]
where \( \mu_0 \) is the permeability of free space, approximately \( 4 \pi \times 10^{-7} \, \text{T m/A} \).

### Step 3: Calculate the magnetic field due to wire 1
Substituting the values into the formula for \( B \):
\[ B = \frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.1} \]
Simplifying this:
\[ B = \frac{4 \times 10^{-7} \times 5}{0.2} = \frac{20 \times 10^{-7}}{0.2} = 100 \times 10^{-7} = 1 \times 10^{-5} \, \text{T} \]

### Step 4: Calculate the force on wire 2
The force \( F \) on a length \( L \) of wire 2 due to the magnetic field created by wire 1 is given by:
\[ F = I_2 \times L \times B \]
Substituting \( I_2 = 5 \, \text{A} \) and \( B = 1 \times 10^{-5} \, \text{T} \):
\[ F = 5 \times L \times 1 \times 10^{-5} \]

### Step 5: Calculate the force per unit length
To find the force per unit length \( \frac{F}{L} \):
\[ \frac{F}{L} = 5 \times 1 \times 10^{-5} = 5 \times 10^{-5} \, \text{N/m} \]

### Conclusion
The force per unit length between the two wires is:
\[ \frac{F}{L} = 5 \times 10^{-5} \, \text{N/m} \]