91Ó°ÊÓ

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:

$F=\frac{{\mathrm{μ}}_0{i}_1{i}_2}{\text{2}\mathrm{π}r}$

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,

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

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,

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

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,

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

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,

$$\begin{gathered} tan\mathrm{θ}=\frac{F_C}{F_B} \\ \text{tan}\mathrm{θ}=\left(\frac{\frac{{\mathrm{μ}}_0{i}^2}{\text{2}\mathrm{π}D}}{\frac{{\mathrm{μ}}_0{i}^2}{\text{2}\mathrm{π}d}}\right) \\ \text{tan}\mathrm{θ}=\frac{d}{D} \end{gathered}$$

From the given figure,

$d<D$

That is,

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

Hence,

$\text{tan}\mathrm{θ}<\text{1}$

This implies that,

role="math" localid="1663004728617" $\mathrm{θ}<{45}^{∘}$

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

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.