Figure 29-81 shows a wire segment of length $\mathit{\Delta }\mathit{s}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{\text{cm}}$, centered at the origin, carrying current $\mathit{i}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\text{A}}$ in the positive ydirection (as part of some complete circuit). To calculate the magnitude of the magnetic field produced by the segment at a point several meters from the origin, we can use $\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{i}\mathbf{\Delta }\overrightarrow{\mathbf{s}}\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$ as the Biot–Savart law. This is because r and u are essentially constant over the segment. Calculate (in unit-vector notation) at the$\left(x,y,z\right)$coordinates (a)localid="1663057128028" $(0,0,5m)$(b)localid="1663057196663" $(0,6m,0)$(c) localid="1663057223833" $\mathbf{(}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{7}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}$and (d)$\left(-3\text{m},-4\text{m},0\right)$

1. Length segment$∆s=3\text{cm}$
2. Current $i=2\text{A}$

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.

Formula:

$\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{i}\overrightarrow{\mathbf{∆}\mathbf{S}}\mathbf{\times }\overrightarrow{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{3}}}$

In the figure, the co-ordinate axis is the center of the cylinder. By symmetry, we will get the same value of magnetic field if we take the cross-sectional area of the left or right side of the cylinder. We take the right side cross-sectional area of the cylinder.

Biot- Savart law can be written as-

$$\begin{gathered} \overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{i\overrightarrow{\Delta s}\times \hat{r}}{r^2} \\ =\frac{{\mu }_0}{4\pi }\frac{i\overrightarrow{\Delta s}\times \overrightarrow{r}}{r^3} \end{gathered}$$

$\overrightarrow{\Delta s}=\Delta s\hat{j}$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{\Delta s}\times \overrightarrow{r}=\Delta s\hat{j}\times \left(x\hat{i}+y\hat{j}+z\hat{k}\right)$

$\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{i}=-\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{j}\times \hat{j}=0$

$\Delta \overrightarrow{s}\times \overrightarrow{r}=\Delta s\left(z\hat{i}-x\hat{k}\right)$

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{i\Delta s\left(z\hat{i}-x\hat{k}\right)}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}$

The magnetic field at the point: $\left(0,0,5m\right)$

Here$\left(x=0,y=0,z=5\text{m}\right)$

Substituting in 1) we get,

$$\begin{gathered} \overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(2\text{A}\right)\left(3\times {10}^{-2}\text{m}\right)\left(5\hat{i}-0\hat{k}\right)\text{m}}{4\pi {(0^2+0^2+{\left(5\text{m}\right)}^2)}^{\frac{3}{2}}} \\ \overrightarrow{B}=\left(2.4\times {10}^{-10}\text{T}\right)\hat{i} \end{gathered}$$

The magnetic field at the point: $\left(0,6\text{m},0\right)$

Here $x=0,y=6\text{m},z=0$

$$\begin{gathered} \overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(2A\right)\left(3\times {10}^{-2}\text{m}\right)\left(0\hat{i}-0\hat{k}\right)\text{m}}{4\pi {(0^2+6^2+0^2)}^{\frac{3}{2}}{\text{m}}^{\text{3}}} \\ \overrightarrow{B}=0 \end{gathered}$$

The magnetic field at the point $\left(7m,7m,0\right)$:

Herelocalid="1663061755332" $x=7\text{m},y=7\text{m},z=0$

localid="1663060711200" $$\begin{gathered} \overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(2A\right)\left(3\times {10}^{-2}\text{m}\right)\left(0\hat{i}-7\hat{k}\right)\text{m}}{4\pi {(7^2+7^2+0^2)}^{\frac{\text{3}}{\text{2}}}{\text{m}}^{\text{3}}} \\ \overrightarrow{B}=\left(-4.3\times {10}^{-11}\text{T}\right)\hat{\mathrm{k}} \end{gathered}$$

The magnetic field at the point $\left(-3\text{m},-4\text{m},0\right)$:

Here , $x=-3 \text{m},y=-4\text{m},z=0$

$$\begin{gathered} \overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(2\text{A}\right)\left(3\times {10}^{-2}\text{m}\right)\left(0\hat{i}+3\hat{k}\right)\text{m}}{4\pi [{\left(-3\right)}^2+{\left(-4\right)}^2+0^2){]}^{\frac{3}{2}}{\text{m}}^{\text{3}}} \\ \overrightarrow{B}=\left(1.44\times {10}^{-10}\text{T}\right)\hat{k} \end{gathered}$$