Question

The two$10\mathrm{cm}$long parallel wires in FIGURE EX29.34 are separated by $5.0\mathrm{mm}$. For what value of the resistor R will the force between the two wires be $5.4\times {10}^{-5}\mathrm{N}$?

Short answer

The value of R is$3\Omega .$

Step-by-step solution

Given information

We have given,

Length of the wire =$10\mathrm{cm}$

Separation between the wire = $5\mathrm{mm}$

Force =$5.4\times {10}^{-5}\mathrm{N}$

We have to find the resistance R as shown in the figure.

Simplify

We know the force between two parallel current carrying wire is

$\mathrm{F}=\frac{{\mathrm{\mu }}_0{\mathrm{I}}_1{\mathrm{I}}_2}{2\mathrm{\pi d}}\mathrm{L}$

using ohm's law in the left circuit,

we can write,

$$\begin{gathered} {\mathrm{I}}_1{\mathrm{R}}_1={\mathrm{V}}_1 \\ {\mathrm{I}}_1=\frac{{\mathrm{V}}_1}{{\mathrm{R}}_1}=\frac{9\mathrm{V}}{2\Omega }=4.5\mathrm{A} \end{gathered}$$

Now substitute the all values that is given in this formula.

$$\begin{gathered} \mathrm{F}=5.4\times {10}^{-5}\mathrm{N}=\frac{(4\mathrm{\pi }\times {10}^{-7})\times 4.5\mathrm{A}\times {\mathrm{I}}_2}{2\mathrm{\pi }\times 0.005} \\ {\mathrm{I}}_2=3\mathrm{A} \end{gathered}$$

Now using the ohm law in right loop, we can write,

$$\begin{gathered} R_2=\frac{{\mathrm{V}}_2}{{\mathrm{I}}_2} \\ {\mathrm{R}}_2=\frac{9\mathrm{V}}{3\mathrm{A}}=3\Omega \end{gathered}$$