Step 1: Understanding the Concept:
Two parallel current-carrying conductors exert a magnetic force on each other.
Currents in the same direction attract each other, while currents in opposite directions repel each other.
The magnitude of this force depends directly on the product of the currents and inversely on the distance between them.
Step 2: Key Formula or Approach:
The magnetic force per unit length between two parallel conductors is $f = \frac{\mu_0 I_1 I_2}{2\pi d}$.
For a constant length $L$, the total force is $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$.
This gives the proportionality: $F \propto \frac{I_1 I_2}{d}$.
Step 3: Detailed Explanation:
Initial State:
Currents $I_1, I_2$ are in the {same direction}. Thus, the initial force $F$ is {attractive}.
\[ F = k \frac{I_1 I_2}{d} \]
where $k = \frac{\mu_0 L}{2\pi}$ is a constant.
Final State:
Distance is increased to $d' = 3d$.
Let the new currents be $I_1'$ and $I_2'$.
The new force is $F' = \frac{2}{3}F$ and it is {repulsive}.
Because the force changed from attractive to repulsive, the direction of one of the currents must have been {reversed}.
Now, let's look at the magnitude:
\[ F' = k \frac{I_1' I_2'}{d'} \]
Substitute $F' = \frac{2}{3}F$ and $d' = 3d$:
\[ \frac{2}{3}F = k \frac{I_1' I_2'}{3d} \]
Substitute the expression for $F$ from the initial state:
\[ \frac{2}{3} \left(k \frac{I_1 I_2}{d}\right) = k \frac{I_1' I_2'}{3d} \]
Cancel the common terms $k$ and $d$:
\[ \frac{2}{3} I_1 I_2 = \frac{I_1' I_2'}{3} \]
Multiply both sides by 3:
\[ 2 I_1 I_2 = I_1' I_2' \]
The product of the new magnitudes must be twice the product of the original magnitudes.
Since the question asks for the change in {one} of the currents, if $I_1' = I_1$, then we must have $I_2' = 2 I_2$.
This means the magnitude of one current must become twice its original value.
Combining both deductions: The magnitude must be twice, and the direction must be reversed.
Step 4: Final Answer:
The change is twice, reversed.