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 the other wire, we can follow these steps:

### Step 1: Understand the Setup
We have two infinite parallel wires that are 10 cm apart, each carrying a current of 20 A in the same direction. We need to find the magnetic force on a 1 cm segment of one wire due to the magnetic field created by the other wire.

**Hint:** Remember that the magnetic field created by a long straight wire can be calculated using the formula for the magnetic field around a wire.

### Step 2: Use the Formula for Magnetic Field
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \) is the permeability of free space.

**Hint:** Ensure you convert the distance from cm to meters when substituting values into the formula.

### Step 3: Calculate the Magnetic Field
In our case, the distance \( r \) between the two wires is 10 cm, which is 0.1 m. Substituting the values into the formula:
\[ B = \frac{4\pi \times 10^{-7} \times 20}{2 \pi \times 0.1} \]
Simplifying this gives:
\[ B = \frac{4 \times 10^{-7} \times 20}{0.2} = \frac{8 \times 10^{-6}}{0.2} = 4 \times 10^{-5} \, \text{T} \]

**Hint:** Check your calculations to ensure you have correctly simplified the fractions.

### Step 4: Calculate the Force on the Segment
The magnetic force \( F \) on a length \( L \) of wire carrying current \( I \) in a magnetic field \( B \) is given by:
\[ F = BIL \]
Here, \( I = 20 \, \text{A} \) and \( L = 1 \, \text{cm} = 0.01 \, \text{m} \).
Substituting the values:
\[ F = (4 \times 10^{-5}) \times 20 \times 0.01 \]
Calculating this gives:
\[ F = 4 \times 10^{-5} \times 20 \times 0.01 = 8 \times 10^{-6} \, \text{N} \]

**Hint:** Make sure to multiply the values carefully and keep track of the powers of ten.

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