91Ó°ÊÓ

Two current -carrying wires generate forces between them since ordinary currents produce magnetic fields and this fields exert significant forces on ordinary currents. The forces on the wires re equal in magnitude and it is given by equation (29.27) in the form

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

where d is the distance between the two wires and $l$is the length of the wire were both wires have the same length, The force is attractive if the current are in the same direction and repulsive if they are in opposite directions.

We can assume that the current in the left circuit is $I_1$, so we use Ohms law to get this current as Ohm's law states that the current $I$flows through a resistance due to the potential difference $V$between the resistance. $I=\frac{V}{R}$

let us determine the current through the left wire using the values for $R=2\mathrm{Î\copyright }$and $V=9V$of the battery by

$I_1=\frac{V}{R}=\frac{9V}{2\mathrm{Î\copyright }}=4.5A$

We rearrange equation(1) for $I_2$to be in the form

$I_2=\frac{2{\mathrm{Ï€»åF}}_{\mathrm{B}}}{{\mathrm{μ}}_0{I}_{1}l}$

Where $I_2$is the current in right wire. Now, plug the values for $d,F_B,{\mathrm{μ}}_0,I_{1}andl$into equation(3) to get $I_2$

$$\begin{gathered} I_2=\frac{2{\mathrm{Ï€»åF}}_{\mathrm{B}}}{{\mathrm{μ}}_0{I}_{1}l} \\ =\frac{2\mathrm{Ï€}(5×{10}^{-3}\mathrm{m})(5.4×{10}^{-5}\mathrm{N})}{(4\mathrm{Ï€}×{10}^{-7}\mathrm{H}/\mathrm{m})(4.5\mathrm{A})(0.10\mathrm{m})} \\ =3A \end{gathered}$$

Again we use the Ohms law to get the resistance $R$in the right circuit by using the value of the potential difference $V=9Vand{I}_2=3A$.

$R=\frac{∆V}{I_2}=\frac{9V}{3A}=3\mathrm{Î\copyright }.$