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Step-by-Step Solution
Step 1: Recall the original force formula between two parallel currents
The force per unit length between two parallel current-carrying conductors, separated by a distance $d$, carrying currents $I_1$ and $I_2$, is given by:
$F_{\text{original}} = \frac{\mu_0}{4\pi} \times \frac{2 I_1 I_2}{d} \times \ell$
where $\ell$ is the length of the conductors in the region where they run parallel.
Here, let this original force be $F$. Thus,
$F = \frac{\mu_0}{4\pi} \times \frac{2 I_1 I_2}{d} \times \ell$
Step 2: Identify the changes in the new scenario
The current in one conductor is doubled: $I_1$ becomes $2I_1$.
The direction of this doubled current is reversed, which introduces a negative sign in the force (since reversing current reverses the direction of the force).
The distance between the two conductors is tripled: $d$ becomes $3d$.
Step 3: Write the new force expression
Taking the new current in one conductor as $2I_1$ and the same current $I_2$ in the other, and the new separation as $3d$, the new force (per unit length) is:
$F' = \frac{\mu_0}{4\pi} \times \frac{2 \bigl(2I_1\bigr) I_2}{3d} \times \ell$
However, because the current has been reversed in direction, the force changes sign. Hence:
$F' = -\,\frac{\mu_0}{4\pi} \times \frac{2 \bigl(2I_1\bigr) I_2}{3d} \times \ell$
Step 4: Relate the new force to the original force
We factor out the original force expression to compare:
$F = \frac{\mu_0}{4\pi} \times \frac{2 I_1 I_2}{d} \times \ell.$
Then,
$\frac{F'}{F} \;=\; \frac{-\,\frac{\mu_0}{4\pi} \times \frac{2\,(2I_1)\,I_2}{3d} \times \ell}{\frac{\mu_0}{4\pi} \times \frac{2\,I_1\,I_2}{d} \times \ell}
\;=\; \frac{-\,2}{3}.$
Thus,
$F' \;=\; -\frac{2}{3}F.$
Step 5: Conclude the answer
The new force between the two conductors becomes:
$\boxed{- \frac{2F}{3}}.$
The negative sign indicates that the direction of the force is opposite to the original one due to the reversal of current in one conductor.
