The given figure displays a segment of an electric circuit. The...

Since the wires $AB$, $CD$, and $EF$ have identical resistances and are connected in parallel, the total current of $30 \text{ A}$ divides equally among them.

Current in each wire, $I = \dfrac{30}{3} = 10 \text{ A}$.

The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \dfrac{\mu_0 I_1 I_2}{2 \pi r}$. Since the currents are in the same direction, the force is attractive.

For wire $AB$, the forces due to $CD$ and $EF$ are both attractive and act downwards.

$f_{AB} = f_{AB, CD} + f_{AB, EF}$

$f_{AB} = \dfrac{\mu_0 I^2}{2 \pi d} + \dfrac{\mu_0 I^2}{2 \pi (2d)} = \dfrac{3 \mu_0 I^2}{4 \pi d}$

Substituting $I = 10 \text{ A}$, $d = 1.0 \text{ cm} = 0.01 \text{ m}$, and $\mu_0 = 4 \pi \times 10^{-7} \text{ T m A}^{-1}$:

$f_{AB} = \dfrac{3 \times 4 \pi \times 10^{-7} \times (10)^2}{4 \pi \times 0.01} = 3 \times 10^{-3} \text{ N m}^{-1}$ downwards.

For wire $CD$, it experiences an upward attractive force from $AB$ and a downward attractive force from $EF$. Since the currents and distances are equal, these forces are equal and opposite.

$f_{CD} = f_{CD, EF} - f_{CD, AB} = \dfrac{\mu_0 I^2}{2 \pi d} - \dfrac{\mu_0 I^2}{2 \pi d} = 0$.