Two infinite, parallel wires are a distance 10 cm apart and carry equal current 20 A in the same direction. The magnetic force on a 1 cm long segment of one wire due to the other wire is ___________ `xx10^(-6)N`. (Permeabilityi of vacuum `mu_(0)=4pixx10^(-7)N//A^(2))`

Explanation

To solve the problem of finding the magnetic force on a 1 cm long segment of one wire due to another infinite, parallel wire carrying equal currents, we can follow these steps:

### Step 1: Identify the parameters
- Distance between the wires, \( d = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current in each wire, \( I = 20 \, \text{A} \)
- Length of the segment, \( L = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \)

### Step 2: Calculate the magnetic field due to one wire at the location of the other wire
The magnetic field \( B \) created by a long straight wire at a distance \( r \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
Here, \( r \) is the distance between the two wires, which is \( 0.1 \, \text{m} \).

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 20}{2 \pi \times 0.1} \]
Simplifying this:
\[ B = \frac{(4 \times 20 \times 10^{-7})}{2 \times 0.1} = \frac{80 \times 10^{-7}}{0.2} = 400 \times 10^{-7} \, \text{T} = 4 \times 10^{-5} \, \text{T} \]

### Step 3: Calculate the magnetic force on the segment of the wire
The magnetic force \( F \) on a current-carrying wire in a magnetic field is given by:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
Where \( \theta \) is the angle between the current direction and the magnetic field direction. Since the current and magnetic field are perpendicular to each other, \( \sin(90^\circ) = 1 \).

Substituting the values:
\[ F = 20 \, \text{A} \cdot 0.01 \, \text{m} \cdot 4 \times 10^{-5} \, \text{T} \]
Calculating this:
\[ F = 20 \cdot 0.01 \cdot 4 \times 10^{-5} = 0.000008 \, \text{N} = 8 \times 10^{-6} \, \text{N} \]

### Final Answer
Thus, the magnetic force on a 1 cm long segment of one wire due to the other wire is:
\[ \boxed{8 \times 10^{-6} \, \text{N}} \]