91Ó°ÊÓ

1. Current$i=2\mathrm{A}$
2. Half circle of radius $r_1=4\mathrm{m}$
3. Two-quarter circles of radius $r_2=r_3=2\mathrm{m}$

An equation known as the Biot-Savart Law describes the magnetic field produced by a steady electric current. It connects the electric current's strength, direction, length, and proximity to the magnetic field.

We use Biot-Savart’s law to find the magnetic field due to straight sections and using the formula of magnetic field due to circular section at the centre, we can find the magnetic field at the centre.

Formulae:

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}}{\mathbf{4}\mathbf{π}\mathbf{d}}\mathit{ϕ}$

$\mathit{d}\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\left(ids\right)\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}\mathbf{}}{{\mathbf{r}}^{\mathbf{2}}}$

The three radial segments do not contribute to produce the magnetic field at the center. Because the current element and radial part are along the same direction and their cross product is zero, for any current element of length $d\stackrel{→}{s}$the angle $\mathrm{θ}$between $d\stackrel{→}{s}$and $\hat{r}$is zero.

$dB=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\left(ids\right)\text{sin}\mathrm{θ}}{r^2}$

$$\begin{gathered} dB=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\left(ids\right)\text{sin}0}{r^2} \\ =0 \end{gathered}$$

Thus, the current along the entire section of three straight wires contributes no magnetic field at the center.

The magnetic field at the center of circular section is

$B=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}d}\mathrm{ϕ}$

Where is the angle made at the center by the circular arc, and is the radius of the circle.

The magnetic field due to a larger radius of wire1 at the center is

$B_1=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(4\text{m}\right)}\left(\mathrm{π}\text{rad}\right)$

Using the right-hand rule, its direction is along positive $Z$direction that is out of the page.

${\stackrel{→}{B}}_1=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(4\text{m}\right)}\left(\mathrm{π}\text{rad}\right)\hat{k}$

Due to one quarter circle, the magnetic field at the center is

${\stackrel{→}{B}}_2=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(2\text{m}\right)}\left(\mathrm{π}/2\text{rad}\right)\hat{k}$

Another quarter circle gives the magnetic field inside the page along direction.

${\stackrel{→}{B}}_3=-\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(2\text{m}\right)}\left(\frac{\mathrm{π}}{2}\text{rad}\right)\hat{k}$

The net magnetic field at the center is then

${\stackrel{→}{B}}_{net}={\stackrel{→}{B}}_1+{\stackrel{→}{B}}_2+{\stackrel{→}{B}}_3$

role="math" localid="1663006402402" $$\begin{gathered} {\stackrel{→}{B}}_{net}=\left(\frac{{\mathrm{μ}}_0\mathrm{i}}{4\mathrm{π}\left(4\text{m}\right)}\left(\mathrm{π}\text{rad}\right)\hat{\mathrm{k}}\right)+\left(\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(2\text{m}\right)}\left(\mathrm{π}/2\text{rad}\right)\hat{\mathrm{k}}\right)-\left(\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(2\text{m}\right)}\left(\frac{\mathrm{π}}{2}\text{rad}\right)\hat{\mathrm{k}}\right) \\ =\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}\left(4\text{m}\right)}\left(\mathrm{π}\text{rad}\right)\hat{\mathrm{k}} \\ =\frac{{\mathrm{μ}}_{0}i}{16}\hat{\mathrm{k}} \end{gathered}$$

=$\frac{\left(4\mathrm{π}×{10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{T}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}\right)\left(2\mathrm{A}\right)}{16}$

${\stackrel{→}{B}}_{net}=1.575×{10}^{-7}\mathrm{T}$

Magnitude of net magnetic field at the centre is$1.575×{10}^{-7}\text{T}.$