91Ó°ÊÓ

Two current-carrying wires generate forces between since ordinary current produce magnetic fields and these fields exert significant forces on ordinary currents. The forces on the wires are equal in magnitude and its given by equation (29.27) in the form

$F_B=\frac{{\mathrm{μ}}_0{I}_1{I}_2}{2\mathrm{Ï€»å}}l$

Where $d$is the distance between the two wires and $l$is the length of wire. The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.

Wire 1 experiences force by wire 2 and 3 and both have current in opposite directions, so the force that is exerted on wire 1 is the difference between the forces if the both wires

$$\begin{gathered} F_1=\frac{{\mathrm{μ}}_0{I}_1{I}_2}{2{\mathrm{Ï€»å}}_{12}}l-\frac{{\mathrm{μ}}_0{I}_1{I}_3}{2{\mathrm{Ï€»å}}_{13}} \\ =\frac{{\mathrm{μ}}_{0}l}{2\mathrm{Ï€}}\left(\frac{I_1{I}_2}{d_{12}}-\frac{I_1{I}_3}{d_{13}}\right) \end{gathered}$$

Where role="math" localid="1649246662514" $d_{12}=2cm$and $d_{13}=4cm.$Now, we plug the values for these parameters to get get the force on wire 1

$$\begin{gathered} F_1=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}}\left(\frac{I_1{I}_2}{d_{12}}-\frac{I_1{I}_3}{d_{13}}\right) \\ =\frac{{\mathrm{μ}}_0(0.50m)}{2\mathrm{π}}\left(\frac{{(10A)}^2}{0.02m}-\frac{{(10A)}^2}{0.04m}\right) \\ =2.5×{10}^{-4}N \end{gathered}$$

The positive sign indicates that the net force is upward.

Wire 2 experiences force wire by 1 and 3 and both have current in the same direction, so the force that is exerted on wire 2 is the difference of the both wires

$F_2=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}}\left(\frac{I_2{I}_1}{d_{12}}-\frac{I_2{I}_3}{d_{23}}\right)$

Where localid="1649247642224" $d_{12}=2cm$and localid="1650882416742" $d_{23}=2cm$. Now, we plug all the values for these parameters to get the force on wire 2

$$\begin{gathered} F_2=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}}\left(\frac{I_2{I}_1}{d_{12}}-\frac{I_2{I}_2}{d_{23}}\right) \\ =\frac{{\mathrm{μ}}_0(0.50m)}{2\mathrm{π}}\left(\frac{{\left(10A\right)}^2}{0.02m}-\frac{{(10A)}^2}{0.02m}\right) \\ =0N \end{gathered}$$

Wire 3 experiences force by wire 1 and 2 and both have current in opposite directions, so the force that is exerted on wire 3 is the difference between the forces of both wires

$F_3=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}}\left(\frac{I_3{I}_1}{d_{13}}-\frac{I_3{I}_2}{d_{23}}\right)$

where $d_{13}=4cmand{d}_{23}=2cm.$Now, we plug the values for these parameters to get the force on wire 3

$F_3=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}}\left(\frac{I_3{I}_1}{d_{13}}-\frac{I_3{I}_2}{d_{23}}\right)$

localid="1649249397893" $$\begin{gathered} =\frac{{\mathrm{μ}}_0(0.50m)}{2\mathrm{π}}\left(\frac{{(10A)}^2}{0.04m}-\frac{{(10A)}^2}{0.02m}\right) \\ =-2.5×{10}^{-4}N \end{gathered}$$

The negative sign indicates that the net force is downward.