Two straight parallel wires, both carrying `10` ampere in the same direction attract each other with a force of `1xx10^(-3)N`. If both currents are doubled, the force of attraction will be

Options

A. `1xx10^(-3)N`
B. `2xx10^(-3)N`
C. `4xx10^(-3)N`
D. `0.25xx10^(-3)N`

Explanation

To solve the problem, we need to understand how the force between two parallel wires carrying current is related to the currents flowing through them. The force \( F \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) can be expressed as:

\[ F = k \cdot \frac{I_1 \cdot I_2 \cdot L}{d} \]

Where:
- \( F \) is the force between the wires,
- \( k \) is a constant that depends on the magnetic permeability of the medium,
- \( L \) is the length of the wires,
- \( d \) is the distance between the wires.

In this case, we are given:
- Initial currents \( I_1 = I_2 = 10 \, \text{A} \)
- Initial force \( F = 1 \times 10^{-3} \, \text{N} \)

### Step 1: Determine the initial force
The initial force between the wires is given as:
\[ F = 1 \times 10^{-3} \, \text{N} \]

### Step 2: Double the currents
Now, if both currents are doubled:
\[ I_1' = I_2' = 2 \times 10 \, \text{A} = 20 \, \text{A} \]

### Step 3: Calculate the new force
Since the force is proportional to the product of the currents, we can express the new force \( F' \) as:
\[ F' = k \cdot \frac{I_1' \cdot I_2' \cdot L}{d} = k \cdot \frac{(2I_1)(2I_2) \cdot L}{d} = 4 \cdot k \cdot \frac{I_1 \cdot I_2 \cdot L}{d} = 4F \]

### Step 4: Substitute the initial force
Substituting the value of \( F \):
\[ F' = 4 \cdot (1 \times 10^{-3} \, \text{N}) = 4 \times 10^{-3} \, \text{N} \]

### Conclusion
Thus, the new force of attraction when both currents are doubled is:
\[ F' = 4 \times 10^{-3} \, \text{N} \]

### Final Answer
The force of attraction will be \( 4 \times 10^{-3} \, \text{N} \).

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