Question: Figure 29-56ashows two wires, each carrying a current .Wire 1 consists of a circular arc of radius Rand two radial lengths; it carries current ${\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{A}}$in the direction indicated. Wire 2 is long and straight; it carries a current i₂ that can be varied; and it is at distance$\frac{\mathbf{R}}{\mathbf{2}}$from the center of the arc. The net magnetic field$\overrightarrow{\mathbf{B}\mathbf{}}$ due to the two currents is measured at the center of curvature of the arc. Figure 29-56bis a plot of the component of in the direction perpendicular to the figure as a function of current i₂. The horizontal scale is set by${\mathit{i}}_{\mathbf{2}\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{A}}$. What is the angle subtended by the arc?

i) The current flowing through the wire is $i_1=2.00A$.

ii) The horizontal scale is set by $i_{2x}=1.00\text{A}$

ii) The radius of the circular arc is R.

iv) The distance between wire 2 and the center of the circular arc of wire 1 is $\frac{R}{2}$.

Use the concept of the magnetic force due to current in straight wires and current in circular arc.

Formulae:

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

$B_{arc}=\frac{{\mu }_{0}i\varnothing }{4\pi R}$

The angle subtended by the arc:

The magnetic field due to a current in long straight wire is

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

Here, from the figure,

$B_{straight}=\frac{{\mu }_0{i}_2}{2\pi \left(\frac{R}{2}\right)}$

According to the right hand, both give direction of magnetic field pointing out of the page.

The magnetic field due to the current in circular arc of the wire is

$B_{arc}=\frac{{\mu }_0{i}_1\varnothing }{4\pi R}$

In the figure (a), according to the right hand, both wires givethedirection of magnetic field pointing out of the page.

The net magnetic field is

$B=B_{arc}-B_{straight}$

$B=\frac{{\mu }_0{i}_1\varnothing }{4\pi R}-\frac{{\mu }_0{i}_2}{2\pi \left(\frac{R}{2}\right)}$ …… (1)

From the figure (b), for $i_{2x}=0.5\text{A,}$then$B=0\text{T}$

Hence, equation (1) becomes

$0=\frac{{\mu }_0{i}_1\varnothing }{4\pi R}-\frac{{\mu }_0{i}_2}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu }_0{i}_1\varnothing }{4\pi R}=\frac{{\mu }_0{i}_2}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu }_0{i}_1\varnothing }{4\pi R}=\frac{{\mu }_0{i}_2}{2\pi \left(\frac{R}{2}\right)}$

Solve further as:

role="math" localid="1663153369320" $\varnothing =4\left(\frac{i_2}{i_1}\right)$

$\varnothing =4\left(\frac{0.50A}{2.00A}\right)$

$\varnothing =1.00\text{rad}$