91Ó°ÊÓ

The current density J is varying across the cross section, so the current is undefined momentarily. Integration of J over the cross section by dividing the wire into concentric strips of thickness dr and radius r will give the total value of the J.

The differential area is$dA=2\mathrm{Ï€°ù»å°ù}$and the current passing through it is dI.

So,

localid="1668264081826" $$\begin{gathered} dl=JdA \\ =2\mathrm{Ï€}\mathrm{Î\pm }{r}^{2}dr \end{gathered}$$

Therefore, the total current is,

$\begin{array}{r}I=∫dI\\ =2\mathrm{Ï€}\mathrm{Î\pm }{∫}_0^R r^{2}dr\\ =2\mathrm{Ï€}\mathrm{Î\pm }\left(\frac{R^3}{3}\right)\end{array}$

Thus, solve for the constant from above equation,

$\mathrm{Î\pm }=\frac{3l}{2\mathrm{Ï€}{R}^3}$

(i) For r $≤$R

Apply Ampere’s circuital law for the circle of radius r $≤$R

localid="1668264177883" $\begin{array}{r}âˆ\circledR \stackrel{→}{B}â‹\dots d\stackrel{→}{l}=Bâˆ\circledR dl\\ =B\left(2\mathrm{Ï€}r\right)\\ ={\mathrm{μ}}_0{l}_{\text{end}}\end{array}$ ...(i)

The current enclosed within the circle is,

$\begin{array}{r}{I}_{\text{end}}=2\mathrm{Ï€}\mathrm{Î\pm }{∫}_0^r r^{2}dr\\ =\frac{2\mathrm{Ï€}\mathrm{Î\pm }{r}^3}{3}\\ =\frac{âˆ\pounds r^3}{R^3}\end{array}$

Thus, by equation (i) for the magnetic field,

$\begin{array}{r}B\left(2\mathrm{π}r\right)={\mathrm{μ}}_0\left(\frac{I{r}^3}{R^3}\right)\\ B=\frac{{\mathrm{μ}}_0/r^2}{2\mathrm{π}{R}^3}\end{array}$

The total current enclosed in the circle of radiusr$â‰\yen$R is I.

Thus, the magnetic field according to Ampere’s Law is,

$\begin{array}{r}B\left(2\mathrm{π}r\right)={\mathrm{μ}}_{0}l\\ B=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}r}\end{array}$