91Ó°ÊÓ

Figure 29-31.

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:

$F_B=\left(\frac{{\mathrm{μ}}_0{i}_a{i}_b}{2\mathrm{π}d}\right)L$

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,

${\mathrm{μ}}_0=$ permeability of vacuum,

$d=$ distance between two 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,

$F_B=\left(\frac{{\mathrm{μ}}_0{i}_a{i}_b}{2\mathrm{π}d}\right)L$

Consider upward and right direction as positive.

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

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

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

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

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

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

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

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

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.