91Ó°ÊÓ

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:

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\mathbf{i}\stackrel{\mathbf{→}}{\mathbf{∆}\mathbf{S}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{3}}}$

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-

$$\begin{gathered} \stackrel{→}{B}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{i\stackrel{→}{\mathrm{Δ}s}×\hat{r}}{r^2} \\ =\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{i\stackrel{→}{\mathrm{Δ}s}×\stackrel{→}{r}}{r^3} \end{gathered}$$

$\stackrel{→}{\mathrm{Δ}s}=\mathrm{Δ}s\hat{j}$

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

$\stackrel{→}{\mathrm{Δ}s}×\stackrel{→}{r}=\mathrm{Δ}s\hat{j}×\left(x\hat{i}+y\hat{j}+z\hat{k}\right)$

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

$\mathrm{Δ}\stackrel{→}{s}×\stackrel{→}{r}=\mathrm{Δ}s\left(z\hat{i}-x\hat{k}\right)$

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{i\mathrm{Δ}s\left(z\hat{i}-x\hat{k}\right)}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}$

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,

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

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

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

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

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

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

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

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