1. Force between two long parallel conductors

For two long parallel wires separated by distance $r$, carrying currents $I_1$ and $I_2$, the magnitude of force is

$F = \frac{\mu_0 I_1 I_2 \ell}{2\pi r}$

where $\ell$ is the common length considered.

Also:

• same direction currents $\Rightarrow$ attractive force
• opposite direction currents $\Rightarrow$ repulsive force

2. Initial force

Initially, distance between wires is $d$, so

$F = \frac{\mu_0 I_1 I_2 \ell}{2\pi d}$

This is given as the original force $F$.

3. New situation

Now:

• current in one wire is doubled and reversed
• distance becomes $3d$

Let $I_1 \to -2I_1$ and $I_2$ remain unchanged.

Then new force:

$F' = \frac{\mu_0 (-2I_1) I_2 \ell}{2\pi (3d)}$

$F' = -\frac{2}{3}\cdot \frac{\mu_0 I_1 I_2 \ell}{2\pi d}$

But

$\frac{\mu_0 I_1 I_2 \ell}{2\pi d} = F$

So,

$F' = -\frac{2F}{3}$

4. Sign interpretation

The negative sign indicates that the nature of force has changed from attractive to repulsive because one current is reversed.

5. Option check

• A: $-\dfrac{2F}{3}$ ✅
• B: $\dfrac{F}{3}$ ❌
• C: $-2F$ ❌
• D: $-\dfrac{F}{3}$ ❌

Thus, the correct answer is A.