Question

Question: Figure 29-31 shows four arrangements in which long, parallel, equally spaced wires carry equal currents directly into or out of the page. Rank the arrangements according to the magnitude of the net force on the central wire due to the currents in the other wires, greatest first.

Short answer

The ranking of the arrangements according to the magnitude of the net force on the central wire due to currents in the other wires is $\mathrm{b}>\mathrm{d}>\mathrm{c}>\mathrm{a}$.

Step-by-step solution

Step 1: Given

Figure 29-31.

Determining the concept.

The force on the central wire is due to all other wires adding the forces by using the formula for the magnetic force between two wires. Comparing them, they can rank.

The formula are as follows:

$F_B=\left(\frac{{\mu }_0{i}_a{i}_b}{2\pi d}\right)L$

Where,

$i_a=$ current carried by first wire,

$i_b=$ current carried by second wire,

role="math" localid="1663000012767" $F=$ force acting on a wire of length L,

$L=$ length of wire,

${\mu }_0=$ permeability of vacuum,

$d=$ distance between two wires.

Determining the arrangements according to the magnitude of the net force on the central wire due to currents in the other wires.

Let the total length of wires be L and the distance between two wires be d.

The magnetic force between two wires is given by,

$F_B=\left(\frac{{\mu }_0{i}_a{i}_b}{2\pi d}\right)L$

Consider upward and right direction as positive.

Determining the arrangement for (a).

$$\begin{gathered} F_B=\left(+\frac{{\mu }_0{i}^2}{4\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}+\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{4\pi d}\right)L \\ {F}_B=0 \end{gathered}$$

Hence, the arrangement for (a) is $F_B=0$.

Determining the arrangement for (b).

$$\begin{gathered} F_B=\left(-\frac{{\mu }_0{i}^2}{4\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{4\pi d}\right)L \\ {F}_B=-\frac{3}{2}\frac{{\mu }_0{i}^2}{\pi d}L \\ \left|F_B\right|=\frac{3}{2}\frac{{\mu }_0{i}^2}{\pi d}L=1.5\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L \end{gathered}$$

Hence, the arrangement for (b) is $\left|F_B\right|=1.5\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L$.

Determining the arrangement for (c).

$$\begin{gathered} F_B=\left(+\frac{{\mu }_0{i}^2}{4\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}4\frac{{\mu }_0{i}^2}{4\pi d}\right)L \\ {F}_B=-\frac{1}{2}\frac{{\mu }_0{i}^2}{\pi d}L \\ \left|F_B\right|=\left(\frac{1}{2}\frac{{\mu }_0{i}^2}{\pi d}\right)L=0.5\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L \end{gathered}$$

Hence, the arrangement for (c) is $\left|F_B\right|=0.5\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L$.

Determining the arrangement for (d).

$$\begin{gathered} F_B=\left(+\frac{{\mu }_0{i}^2}{4\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{2\pi d}-\frac{{\mu }_0{i}^2}{4\pi d}\right)L \\ {F}_B=\left(-\frac{5}{3}\frac{{\mu }_0{i}^2}{\pi d}\right)L \\ \left|F_B\right|=\left(\frac{5}{4}\frac{{\mu }_0{i}^2}{\pi d}\right)L=1.25\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L \end{gathered}$$

Hence, the arrangement for (d) is $\left|F_B\right|=1.25\left(\frac{{\mu }_0{i}^2}{\pi d}\right)L$.

Hence, the ranking of the arrangements according to the magnitude of the net force on the central wire due to currents in the other wires is $\mathrm{b}>\mathrm{d}>\mathrm{c}>\mathrm{a}$.

Magnetic force on a wire due to other wires can be found by using the formula for the magnetic force between two wires.