Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is

Options

A. `5xx10^(-4)N," (attraction)"`
B. `1xx10^(-4)N," (attraction)"`
C. `5xx10^(-4)N," (repulsion)"`
D. `1xx10^(-4)N," (repulsion)"`

Answer

To solve the problem of finding the force acting on two long parallel wires carrying currents in the same direction, we can follow these steps:

### Step 1: Identify the given values
- Distance between the wires, \( d = 30 \, \text{cm} = 0.3 \, \text{m} \)
- Current in wire A, \( I_A = 10 \, \text{A} \)
- Current in wire B, \( I_B = 15 \, \text{A} \)
- Length of the wires considered, \( L = 5 \, \text{m} \)

### Step 2: Use the formula for the force per unit length between two parallel wires
The formula for the force per unit length \( F/L \) between two parallel wires carrying currents \( I_A \) and \( I_B \) is given by:
\[ \frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{I_A I_B}{d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (the permeability of free space).

### Step 3: Substitute the values into the formula
Substituting the values into the formula:
\[ \frac{F}{L} = \frac{(4\pi \times 10^{-7})}{2\pi} \cdot \frac{10 \times 15}{0.3} \]
This simplifies to:
\[ \frac{F}{L} = \frac{2 \times 10^{-7}}{0.3} \cdot 150 \]
Calculating this gives:
\[ \frac{F}{L} = \frac{3 \times 10^{-5}}{0.3} = 1 \times 10^{-4} \, \text{N/m} \]

### Step 4: Calculate the total force over the length of the wires
Now, to find the total force \( F \) acting over the length \( L \):
\[ F = \left(\frac{F}{L}\right) \cdot L = (1 \times 10^{-4}) \cdot 5 = 5 \times 10^{-4} \, \text{N} \]

### Step 5: Determine the nature of the force
Since both currents are in the same direction, the force between the two wires will be attractive.

### Final Answer
The force acting over a length of 5 m of the wires is:
\[ F = 5 \times 10^{-4} \, \text{N} \quad \text{(attractive)} \]

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