The force of rcpulsion between two parallel wires is f when each one of them carries a certain current I. If the current in each is doubled, the force between them would be

Options

A. 4/f
B. 4f
C. 2f
D. f

Explanation

To solve the problem, we need to understand how the force of repulsion between two parallel wires changes when the current in each wire is doubled. The force of repulsion \( F \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) is given by the formula:

\[ F = \frac{\mu_0 I_1 I_2 L}{2 \pi d} \]

where:
- \( F \) is the force between the wires,
- \( \mu_0 \) is the permeability of free space,
- \( I_1 \) and \( I_2 \) are the currents in the two wires,
- \( L \) is the length of the wires,
- \( d \) is the distance between the wires.

### Step-by-Step Solution:

1. **Initial Force Calculation**:
- Let the initial current in each wire be \( I \).
- The initial force \( F \) can be expressed as:
\[ F = \frac{\mu_0 I \cdot I \cdot L}{2 \pi d} = \frac{\mu_0 I^2 L}{2 \pi d} \]

2. **Doubling the Current**:
- If the current in each wire is doubled, then the new currents \( I_1 \) and \( I_2 \) become \( 2I \).
- We need to calculate the new force \( F' \) when both currents are \( 2I \):
\[ F' = \frac{\mu_0 (2I)(2I) L}{2 \pi d} \]

3. **Simplifying the New Force**:
- Substitute \( 2I \) into the equation:
\[ F' = \frac{\mu_0 (4I^2) L}{2 \pi d} \]
- Thus, we can rewrite \( F' \) as:
\[ F' = 4 \cdot \frac{\mu_0 I^2 L}{2 \pi d} \]
- Notice that \( \frac{\mu_0 I^2 L}{2 \pi d} \) is the original force \( F \):
\[ F' = 4F \]

4. **Conclusion**:
- Therefore, when the current in each wire is doubled, the new force of repulsion \( F' \) becomes four times the original force \( F \):
\[ F' = 4F \]

### Final Answer:
The force between the wires when the current in each is doubled is \( 4F \).

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