Figure (a) shows, in cross section, two wires that are straight, parallel, and very long. The ratio ${\mathit{i}}_{\mathbf{1}}\mathbf{/}{\mathit{i}}_{\mathbf{2}}$of the current carried by wire 1 to that carried by wire 2 is$\frac{\mathbf{1}}{\mathbf{3}}$. Wire 1 is fixed in place. Wire 2 can be moved along the positive side of the x-axis so as to change the magnetic energy density u_B set up by the two currents at the origin. Figure (b) gives u_B as a function of the position x of wire 2. The curve has an asymptote of${\mathbf{u}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{nJ}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{as}\mathbf{}\mathbf{x}\mathbf{\to }\mathbf{\infty }$, and the horizontal axis scale is set by${\mathbf{x}}_{\mathbf{s}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. What is the value of (a) i₁ and (b) i₂?

i) Ratio between the currents $i_1/i_2=1/3$

ii) Magnetic energy density$u_B=1.96\mathrm{nJ}/{\mathrm{m}}^3\mathrm{at}\mathrm{x}\to \infty$

From the graph and using the given ratio of the currents, we can find the distance of the wire 1 at which both the magnetic fields get cancel out. Then, using the magnetic field at this distance and magnetic field due to first wire, we can find the current in the first wire. Using this current in the given ratio, we can find the current in the second wire.

Formula:

$$\begin{gathered} u_B=B^2/2{\mu }_0 \\ B={\mu }_0/2\pi d \end{gathered}$$

The graph shows the total magnetic energy density is zero at x = 20 cm . This indicates that the currents are in the same directions and the magnetic fields are in the opposite direction. At this point, the magnetic fields are the same.

$$\begin{gathered} B_1=B_2 \\ \frac{{\mu }_0{i}_1}{2\pi d}=\frac{{\mu }_0{i}_2}{2\pi \left(0.20m\right)} \\ d=\left(0.20m\right)\frac{i_1}{i_2} \\ d=\left(0.20m\right)\frac{1}{3} \\ d=0.0667m \end{gathered}$$

We have given at$x\to \infty$magnetic field due to second wire is zero and therefore, the energy density is due to first wire.

$$\begin{gathered} u_B=\frac{B_1^2}{2{\mu }_0} \\ {B}_1=\sqrt{2{\mu }_0{u}_B} \end{gathered}$$

And the magnetic field due to the first wire is given by,

role="math" localid="1661855343385" $B_1=\frac{{\mu }_0{i}_1}{2\pi d}$

Therefore,

$$\begin{gathered} \frac{{\mu }_0{i}_1}{2\pi d}=\sqrt{2{\mu }_0{u}_0} \\ {i}_1=\left(\frac{2\pi d}{{\mu }_0}\right)\sqrt{2{\mu }_0{u}_B} \\ {i}_1=\left(2\pi d\right)\sqrt{\frac{2u_B}{{\mu }_0}} \\ {i}_1=\left[2\pi \left(0.0667m\right)\right]\sqrt{\frac{2\left(1.96\times {10}^{-9}{\displaystyle \frac{J}{m^3}}\right)}{\left(\pi \times {10}^{-7}{\displaystyle \frac{H}{m}}\right)}} \\ {i}_1=23.4\times {10}^{-3}A \\ {i}_1=23.4mA \end{gathered}$$

Therefore, the value of $i_1\mathrm{is}23.4mA$.

We have the ratio of the currents.

$$\begin{gathered} \frac{i_1}{i_2}=\frac{1}{3} \\ {i}_2=3i_1 \\ {i}_2=\left(23.4mA\right) \\ {I}_2=70.2mA=70A \end{gathered}$$

Therefore, the value of$I_2\mathrm{is}70.2mA$ .