91Ó°ÊÓ

The total current on the wire is $I$and the current density is given by$J=\frac{J_{0}r}{R}$.

Let's consider an Amperian loop of thickness $dr$at a distance $r$from the center of the wire.

The formula to calculate the infinitesimal current $\left(dI\right)$carried by the Amperian loop is given by

$dI=2\mathrm{Ï€}rdrJ...................\left(1\right)$

Substitute the expression for the current density into equation (1) and simplify to obtain the infinitesimal current.

role="math" localid="1649567717430" $\begin{array}{rcl}dI& =& 2\mathrm{Ï€}rdr\left(\frac{J_{0}r}{R}\right)\\ & =& \frac{2\mathrm{Ï€}{J}_0}{R}{r}^{2}dr............................\left(2\right)\end{array}$

The formula to calculate the total current $\left(I\right)$carrying by the wire is given by

$I={∫}_0^{R}dI....................\left(3\right)$

Substitute the expression for $dI$from equation (2) into equation (3) and simplify to obtain the required constant.

role="math" localid="1649567835880" $\begin{array}{rcl}I& =& {∫}_0^R\frac{2\mathrm{π}{J}_0}{R}{r}^{2}dr\\ & =& \frac{2\mathrm{π}{J}_0}{R}{∫}_0^R{r}^{2}dr\\ & =& \frac{2\mathrm{π}{R}^2{J}_0}{3}\\ J_0& =& \frac{3}{2\mathrm{π}{R}^2}I............................\left(4\right)\end{array}$

The required constant is given by$\begin{array}{rcl}{J}_0& =& \frac{3}{2\mathrm{Ï€}{R}^2}I\end{array}$.

The potion of the wire inside it is given whose radius is$r$.

The current $\left(I_r\right)$carrying by the wire of radius $r$can be calculated as before following the calculation shown below.

$\begin{array}{rcl}{I}_r& =& {∫}_0^r\frac{2\mathrm{π}{J}_0}{R}{r}^{2}dr\\ & =& \frac{2\mathrm{π}{r}^3{J}_0}{3R}.....................\left(5\right)\end{array}$

Substitute $\left(\begin{array}{rcl}& & \frac{3}{2\mathrm{Ï€}{R}^2}I\end{array}\right)$for $J_0$into equation (5) to express the enclosed current in terms of the total current.

$\begin{array}{rcl}{I}_r& =& \frac{2\mathrm{Ï€}{r}^3}{3R}\frac{3}{2\mathrm{Ï€}{R}^2}I\\ & =& \frac{r^3}{R^3}I..........................\left(6\right)\end{array}$

Applying Ampere's circuital law for the Amperian loop, it can be written that

role="math" localid="1649568694248" $âˆ\circledR \stackrel{â‡\P Ä}{B}.d\stackrel{â‡\P Ä}{l}={\mathrm{μ}}_0{I}_r...............\left(7\right)$

Here, $B$is the required magnetic field.

Substitute the expression for $I_r$from equation (6) into equation (7) and simplify to obtain the required magnetic field.

role="math" localid="1649568746285" $\begin{array}{rcl}âˆ\circledR \stackrel{â‡\P Ä}{B}.d\stackrel{â‡\P Ä}{l}& =& {\mathrm{μ}}_0\frac{r^3}{R^3}I\\ B\left(2\mathrm{Ï€}r\right)& =& {\mathrm{μ}}_0\frac{r^3}{R^3}I\\ B& =& \frac{{\mathrm{μ}}_{0}I{r}^2}{2\mathrm{Ï€}{R}^3}.........................\left(8\right)\end{array}$

The required magnetic field is given by$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_{0}I{r}^2}{2\mathrm{π}{R}^3}\end{array}$.

Substitute $R$for $r$into equation (8) and simplify to obtain the magnetic field at the boundary of the wire.

$\begin{array}{rcl}{B}_R& =& \frac{{\mathrm{μ}}_{0}I{R}^2}{2\mathrm{π}{R}^3}\\ & =& \frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}R}\end{array}$

As can be seen from the equation above, the results exactly matches the correct answer that one should expect at the boundary.

Yes, the solution matches with the known solution outside a long straight current carrying wire.