Two long straight wires are arranged parallel to each other and are kept 10 cm apart in vacuum. They carry currents of 5A and 10A respectively in the same direction. What is the magnetic force on a lergth of 20 cm of either wire ?

Options

A. `10^(-5)N`
B. `2xx10^(-5)M`
C. `3xx10^(-5)N`
D. `4xx10^(-5)N`

Explanation

To solve the problem of finding the magnetic force between two parallel wires carrying currents, we can follow these steps:

### Step 1: Understand the Problem
We have two long straight wires that are parallel to each other, separated by a distance of 10 cm, carrying currents of 5 A and 10 A in the same direction. We need to find the magnetic force acting on a length of 20 cm of either wire.

### Step 2: Use the Formula for Magnetic Force
The magnetic force per unit length between two parallel wires carrying currents \( I_1 \) and \( I_2 \) is given by the formula:
\[ F/L = \frac{\mu_0 I_1 I_2}{2\pi r} \]
where:
- \( F \) is the total magnetic force,
- \( L \) is the length of the wire,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( r \) is the distance between the wires.

### Step 3: Substitute the Values
Given:
- \( I_1 = 5 \, \text{A} \)
- \( I_2 = 10 \, \text{A} \)
- \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)

Now, substituting these values into the formula:
\[ F/L = \frac{(4\pi \times 10^{-7}) \times 5 \times 10}{2\pi \times 0.1} \]

### Step 4: Simplify the Equation
The \( \pi \) cancels out:
\[ F/L = \frac{4 \times 10^{-7} \times 5 \times 10}{2 \times 0.1} \]
\[ F/L = \frac{20 \times 10^{-7}}{0.2} \]
\[ F/L = 100 \times 10^{-7} = 10^{-5} \, \text{N/m} \]

### Step 5: Calculate the Total Force
Now, to find the total force \( F \):
\[ F = (F/L) \times L = (10^{-5}) \times 0.2 = 2 \times 10^{-6} \, \text{N} \]

### Step 6: Final Result
Thus, the magnetic force on a length of 20 cm of either wire is:
\[ F = 2 \times 10^{-6} \, \text{N} \]

### Summary
The magnetic force acting on a length of 20 cm of either wire is \( 2 \times 10^{-6} \, \text{N} \).

---