91Ó°ÊÓ

• The radius of the wire$\mathrm{a}=3.1\mathrm{mm}=0.0031\mathrm{m}$
• Current density varies as$\mathrm{J}=\frac{{\mathrm{J}}_0\mathrm{r}}{\mathrm{a}}$

• ${\mathrm{J}}_0=310\mathrm{A}/{\mathrm{m}}^2$

The relation between current and current density is,

$\mathbf{i}\mathbf{=}\mathbf{∫}\mathbf{J}\mathbf{}\mathbf{dA}$

Ampere’s law states that,

$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{â‹\dots }\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{=}{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}$

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is the net current encircled by the loop.

By using the current density equation in Ampere’s law and integrating it with respect to distance r, we can get the general equation for the magnetic field due to the current-carrying cylindrical wire. By using this, we can find the value of the magnetic field at given distances.

According to Ampere’s law,

$\begin{array}{c}âˆ\circledR \stackrel{→}{\mathrm{B}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{s}}={\mathrm{μ}}_0{\mathrm{i}}_{\mathrm{enc}}\\ \mathrm{B}âˆ\circledR \mathrm{ds}={\mathrm{μ}}_0{\mathrm{i}}_{\mathrm{enc}}\end{array}$

Since$âˆ\circledR \mathrm{ds}=\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{circular}\mathrm{path}=2\mathrm{Ï€°ù}$

$∴\mathrm{B}\left(2\mathrm{Ï€°ù}\right)={\mathrm{μ}}_0{\mathrm{i}}_{\mathrm{enc}}$ (i)

Current is given by,

${\mathrm{i}}_{\mathrm{enc}}=∫\mathrm{J}\mathrm{dA}$

Using the given current density $\mathrm{J}=\frac{{\mathrm{J}}_0\mathrm{r}}{\mathrm{a}}$

${\mathrm{i}}_{\mathrm{enc}}=∫\frac{{\mathrm{J}}_0\mathrm{r}}{\mathrm{a}}\mathrm{dA}$

Area of a differential element of the circle is$\mathrm{dA}=2\mathrm{Ï€°ù}\mathrm{dr}$

$\begin{array}{c}{\mathrm{i}}_{\mathrm{enc}}=∫\frac{{\mathrm{J}}_0\mathrm{r}}{\mathrm{a}}2\mathrm{Ï€°ù}\mathrm{dr}\\ =\frac{2{\mathrm{Ï€´³}}_0}{\mathrm{a}}∫{\mathrm{r}}^2\mathrm{dr}\\ =\frac{2{\mathrm{Ï€´³}}_0}{\mathrm{a}}\frac{{\mathrm{r}}^3}{3}\end{array}$

Using this in equation (i),

$\mathrm{B}\left(2\mathrm{Ï€°ù}\right)={\mathrm{μ}}_0\frac{2{\mathrm{Ï€´³}}_0}{\mathrm{a}}\frac{{\mathrm{r}}^3}{3}$

Therefore,

$\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{J}}_0}{3}\frac{{\mathrm{r}}^2}{\mathrm{a}}$ (ii)

At $\mathrm{r}=0$, Equation (ii) becomes,

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{J}}_0}{3}×\frac{0}{\mathrm{a}}\\ =0\end{array}$

Thus, the magnetic field at $\mathrm{r}=0$ is zero.

At$\mathrm{r}=\frac{\mathrm{a}}{2}$,Equation$\left(2\right)$becomes

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{J}}_0}{3}×\frac{{\left(\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{\mathrm{a}}\\ \mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{J}}_0\mathrm{a}}{12}\\ =\frac{4\mathrm{Ï€}×{10}^{-7}×310×0.0031}{12}\\ =\frac{4×3.14×{10}^{-7}×310×0.0031}{12}\\ =0.1×{10}^{-6}\mathrm{T}\\ =0.10\mathrm{μ°Õ}\end{array}$

Thus, The magnetic field at $\mathrm{r}=\frac{\mathrm{a}}{2}$ is $\mathrm{B}=0.10\mathrm{μ°Õ}$

At$\mathrm{r}=\mathrm{a}$,Equation$\left(2\right)$becomes

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{J}}_0}{3}×\frac{{\mathrm{a}}^2}{\mathrm{a}}\\ =\frac{{\mathrm{μ}}_0{\mathrm{J}}_0\mathrm{a}}{3}\\ =\frac{4\mathrm{Ï€}×{10}^{-7}×310×0.0031}{3}\\ =\frac{4×3.14×{10}^{-7}×310×0.0031}{3}\\ =4.02×{10}^{-7}\mathrm{T}\\ =0.40\mathrm{μ°Õ}\end{array}$

Thus, the magnetic field at $\mathrm{r}=\mathrm{a}$ is $\mathrm{B}=0.40\mathrm{μ°Õ}$.