Figure 29-34 shows three arrangements of three long straight wires carrying equal currents directly into or out of the page. (a) Rank the arrangements according to the magnitude of the net force on wire Adue to the currents in the other wires, greatest first. (b) In arrangement 3, is the angle between the net force on wire A and the dashed line equal to, less than, or more than 45°?

Figure showing three arrangements in which three long straight wires carry equal currents directly into or out of the page.

Find the directions of forces on A due to other wires using the right-hand rule. Then using the formula for the force between two wires, rank the arrangements according to the magnitude of the net force on wire A due to the currents in the other wires and find the angle between the net force on wire A and the dashed line in arrangement 3.

The formula is as follows:

$F=\frac{{\mu }_0{i}_1{i}_2}{\text{2}\pi r}$

Where,

i₁ = current carried by first wire,

i₂ = current carried by second wire,

F = force acting on a wire,

µ₀ = permeability of vacuum,

d = distance between two wires $\left(r\right)$.

Let, the wire at distance d from A be B and at distance D be C, and the current through A, B, and C be i.

Let’s take the left as positive.

Applying the right-hand rule to the arrangement 1 gives that F_B and F_C are directing left.

Hence, the net magnetic force on wire A is,

$$\begin{gathered} F=F_B+F_C \\ F=\frac{{\mu }_0{i}^2}{\text{2}\pi d}+\frac{{\mu }_0{i}^2}{\text{2}\pi D} \end{gathered}$$

Applying the right-hand rule to the arrangement 2 gives that, F_Cis directing left and F_B is directing right.

Hence, the net magnetic force on wire A is,

$$\begin{gathered} F=F_B-F_C \\ F=\frac{{\mu }_0{i}^2}{\text{2}\pi d}-\frac{{\mu }_0{i}^2}{\text{2}\pi D} \end{gathered}$$

Applying the right-hand rule to arrangement 3 gives that, F_Cis directing up and F_B is directing right.

Hence, the net magnetic force on wire A is,

$$\begin{gathered} F=\sqrt{F_B^2+F_C^2} \\ F=\sqrt{{\left(\frac{{\mu }_0{i}^2}{\text{2}\pi d}\right)}^2+{\left(\frac{{\mu }_0{i}^2}{\text{2}\pi D}\right)}^2} \end{gathered}$$

Hence, the ranking of arrangements according to the magnitude of the net force on wire A due to the currents in the other wires is $\text{1 > 3 > 2}$.

The angle between the net force on wire A and the dashed line is,

$$\begin{gathered} tan\theta =\frac{F_C}{F_B} \\ \text{tan}\theta =\left(\frac{\frac{{\mu }_0{i}^2}{\text{2}\pi D}}{\frac{{\mu }_0{i}^2}{\text{2}\pi d}}\right) \\ \text{tan}\theta =\frac{d}{D} \end{gathered}$$

From the given figure,

$d<D$

That is,

$\frac{d}{D}<\text{1}$

Hence,

$\text{tan}\theta <\text{1}$

This implies that,

role="math" localid="1663004728617" $\theta <{45}^{\circ }$

Hence, the angle between the net force on wire A and the dashed line in arrangement 3 is less than${45}^{\circ }$

Therefore, rank the arrangements according to the magnitude of the net force acting on the wire due to other wires using the right-hand rule and the formula for the force between two wires.