91Ó°ÊÓ

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:

$B_{straight}=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

$B_{arc}=\frac{{\mathrm{μ}}_{0}iâˆ\dots }{4\mathrm{Ï€}R}$

The angle subtended by the arc:

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

$B_{straight}=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

Here, from the figure,

$B_{straight}=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}\left(\frac{R}{2}\right)}$

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

$B_{arc}=\frac{{\mathrm{μ}}_0{i}_1âˆ\dots }{4\mathrm{Ï€}R}$

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

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

$B=\frac{{\mathrm{μ}}_0{i}_1âˆ\dots }{4\mathrm{Ï€}R}-\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{Ï€}\left(\frac{R}{2}\right)}$ …… (1)

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

Hence, equation (1) becomes

$0=\frac{{\mathrm{μ}}_0{i}_1âˆ\dots }{4\mathrm{Ï€}R}-\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{Ï€}\left(\frac{R}{2}\right)}$

$\frac{{\mathrm{μ}}_0{i}_1âˆ\dots }{4\mathrm{Ï€}R}=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{Ï€}\left(\frac{R}{2}\right)}$

$\frac{{\mathrm{μ}}_0{i}_1âˆ\dots }{4\mathrm{Ï€}R}=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{Ï€}\left(\frac{R}{2}\right)}$

Solve further as:

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

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

$âˆ\dots =1.00\text{rad}$