In Fig. 29-54a, wire 1 consists of a circular arc and two radial lengths; it carries current${\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathit{A}$in the direction indicated. Wire 2, shown in cross section, is long, straight, and Perpendicular to the plane of the figure. Its distance from the center of the arc is equal to the radius $\mathit{R}$of the arc, and it carries a current i₂ that can be varied. The two currents set up a net magnetic field$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$at the center of the arc. Figure bgives the square of the field’s magnitude B² plotted versus the square ofthe current${\mathit{i}}_{\mathbf{2}}^{\mathbf{2}}$. The vertical scale is set by${\mathit{B}}_{\mathbf{s}}^{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{}{\mathbf{\text{T}}}^{\mathbf{\text{2}}}$what angle is subtended by the arc?

The current is $i_1=0.50A$.

The vertical scale,${B_s}^2=10\times {10}^{-10}{\text{T}}^{\text{2}}$

Magnetic field at the center of a circular arc is

$B=\frac{{\mu }_{0}i\varphi }{4\pi R}$

Magnetic field due to a long straight wire carrying a current is

$B=\frac{{\mu }_{0}i}{2\pi R}$

The magnitude of the magnetic field at the center of a circular arc of radius Rand central angle $\theta$(in radians), carrying current $i$is given by equation $29-9$as:

$B_1=\frac{{\mu }_0{i}_1\varphi }{4\pi R}$ ……. (1)

For a long straight wire carrying a current $i$, the magnitude of the magnetic field at a perpendicular distance Rfrom the wire is given by equation $29-4$as

$B_2=\frac{{\mu }_0{i}_2}{2\pi R}$ ……. (2)

Using the right-hand rule, we can conclude that the curved section will result in a magnetic field out of the page. Also, the magnetic field due to the second wire is perpendicular to the magnetic field due to the first wire.

Find the net magnetic field at the center of the arc by using Pythagoras theorem.

$B^2={B_1}^2+{B_2}^2$

Thus from equation (1) and (2),

$B^2={\left(\frac{{\mu }_0{i}_1\varphi }{4\pi R}\right)}^2+{\left(\frac{{\mu }_0{i}_2}{2\pi R}\right)}^2$ ……. (3)

The term ${\left(\frac{{\mu }_0}{2\pi R}\right)}^2$gives us the slope of the graph given in $Fig.29-54\left(b\right)$, i.e.,

$$\begin{gathered} \frac{B^2}{{i_2}^2}={\left(\frac{{\mu }_0}{2\pi R}\right)}^2 \\ =\frac{\left(11\times {10}^{-10}-1\times {10}^{-10}\right)}{2-0} \\ =\frac{10\times {10}^{-10}}{2} \\ =5\times {10}^{-10} \end{gathered}$$

Rearranging the equation for $R$

$$\begin{gathered} R^2={\left(\frac{{\mu }_0}{2\pi }\right)}^2\frac{1}{5\times {10}^{-10}} \\ ={\left(\frac{4\pi \times {10}^{-7}}{2\pi }\right)}^2\frac{1}{5\times {10}^{-10}} \\ =8\times {10}^{-5}{\text{m}}^{\text{2}} \end{gathered}$$

$R=8.94\text{mm}$

If substitute$i_2=0$in equation$\left(3\right),$get the y-intercept in the graph.

From the graph, the y-intercept is$1\times {10}^{-10}$

Thus, for$i_2=0$,$B^2=1\times {10}^{-10}$

Therefore, from equation$\left(3\right),$

$B^2={\left(\frac{{\mu }_0{i}_1\varphi }{4\pi R}\right)}^2$

$1\times {10}^{-10}={\left(\frac{{\mu }_0{i}_1\varphi }{4\pi R}\right)}^2$

Rearranging the equation for,${\varphi }^2$

${\varphi }^2=\frac{{\left(4\pi R\right)}^2\left(1\times {10}^{-10}\right)}{{\left({\mu }_0{i}_1\right)}^2}$

Substituting the values and solve as:

${\varphi }^2=\frac{{\left(4\pi \right)}^2{\left(8.94\times {10}^{-3}\text{m}\right)}^2\left(1\times {10}^{-10}\right)}{{\left(\left(4\pi \times {10}^{-7}\text{T}·\frac{\text{m}}{\text{A}}\right)0.50\text{A}\right)}^2}$

${\varphi }^2=3.19$

$$\begin{gathered} \varphi =1.79rad \\ \varphi \approx 1.8rad \end{gathered}$$