Two parallel wires in free spaces are `10cm` apart and each carries a current of `10A` in the same direction. The force one wire exerts on the other per metre of length is

Options

A. `2xx10^(-4)N` , attractive
B. `2xx10^(-4)N` ,repulsive
C. `2xx10^(-7)N` ,attractive
D. `2xx10^(-7)N` ,repulsive

Explanation

To solve the problem of finding the force one wire exerts on the other per meter of length, we can follow these steps:

### Step 1: Identify the Given Values
- Distance between the wires (R) = 10 cm = 0.1 m
- Current in each wire (I1 and I2) = 10 A

### Step 2: Use the Formula for the Magnetic Field
The magnetic field (B) created by one wire at the location of the other wire is given by the formula:
\[ B = \frac{\mu_0 I_1}{2 \pi R} \]
where \(\mu_0\) (the permeability of free space) is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

### Step 3: Calculate the Magnetic Field (B)
Substituting the values into the formula:
\[ B = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 10 \, \text{A}}{2 \pi \times 0.1 \, \text{m}} \]

### Step 4: Simplify the Expression
- The \(\pi\) cancels out:
\[ B = \frac{(4 \times 10^{-7} \times 10)}{2 \times 0.1} \]
- This simplifies to:
\[ B = \frac{4 \times 10^{-6}}{0.2} = 2 \times 10^{-5} \, \text{T} \]

### Step 5: Calculate the Force on the Second Wire
The force (F) on the second wire due to the magnetic field created by the first wire is given by:
\[ F = I_2 \cdot L \cdot B \]
where \(L\) is the length of the wire (1 meter in this case).

Substituting the values:
\[ F = 10 \, \text{A} \cdot 1 \, \text{m} \cdot 2 \times 10^{-5} \, \text{T} \]

### Step 6: Calculate the Force
\[ F = 10 \cdot 2 \times 10^{-5} = 2 \times 10^{-4} \, \text{N} \]

### Step 7: Determine the Direction of the Force
Since both currents are in the same direction, the force between the wires will be attractive.

### Final Answer
The force one wire exerts on the other per meter of length is:
\[ F = 2 \times 10^{-4} \, \text{N} \text{ (attractive)} \]

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