91Ó°ÊÓ

Refer to the image below,

Here, the cross section of the cylinder is divided into thin concentric rings of radius r and dr with current through each ring as dI,

The differential area is $dA=2\mathrm{Ï€}rdr$

So,

$$\begin{gathered} dl=JdA \\ =2\mathrm{Ï€°ù»å°ù} \end{gathered}$$

Integrate the above equation by substituting the current density expression from the question,

$$\begin{gathered} dI=\frac{2I_0}{\mathrm{π}{\mathrm{a}}^2}\left[1−{\left(\frac{r}{a}\right)}^2\right]2\mathrm{π}rdr \\ =\frac{4I_0}{a^2}\left[1−{\left(\frac{r}{a}\right)}^2\right]rdr \end{gathered}$$

Therefore,

$$\begin{gathered} I={∫}_0^a dI \\ =\frac{4I_0}{a^2}{\left[\frac{1}{2}{r}^2−\frac{1}{4}\frac{r^4}{a^2}\right]}_0^a \\ =I_0 \end{gathered}$$

Hence proved.

With the help of Ampere’s Law, a ampere’s circle is drawn as shown below.

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

$$\begin{gathered} \boxed{∫}\stackrel{→}{B}â‹\dots d\stackrel{→}{l}=B\boxed{∫}dl \\ =B\left(2\mathrm{Ï€}r\right) \\ ={\mathrm{μ}}_0{I}_{\text{end}} \\ ={\mathrm{μ}}_0{I}_0 \end{gathered}$$

Thus, the expression of magnetic field islocalid="1668342539849" $B={\mathrm{μ}}_0{I}_0$

Again dividing the cross section into thin concentric strips of radius r' and width dr' as shown below,

Similar to part (a), the current passing through the element chosen is,

$dI=\frac{4I_0}{a^2}\left[1−{\left(\frac{r^{\mathrm{′}}}{a}\right)}^2\right]r^{\mathrm{′}}d{r}^{\mathrm{′}}$

Thus, Integrate the above equation to obtain the expression for current,

$$\begin{gathered} I={∫}_0^a dl \\ =\frac{4I_0}{a^2}{\left[\frac{1}{2}{r}^{\mathrm{′}2}−\frac{1}{4}\frac{r^{\mathrm{′}4}}{a^2}\right]}_0^r \\ =\frac{I_0{r}^2}{a^2}\left(2−\frac{r^2}{a^2}\right) \end{gathered}$$

With the help of Ampere’s Law, a ampere’s circle is drawn as shown below.

Apply Ampere’s circuital law for the circle of radius r $â‰\yen$a

$$\begin{gathered} \boxed{∫}\stackrel{→}{B}â‹\dots d\stackrel{→}{l}=B\boxed{∫}dl \\ =B\left(2\mathrm{Ï€}r\right) \\ ={\mathrm{μ}}_0{I}_{\text{encl}} \\ ={\mathrm{μ}}_0\frac{I_0{r}^2}{a^2}\left(2−\frac{r^2}{a^2}\right) \end{gathered}$$

Thus, the expression of magnetic field is $B={\mathrm{μ}}_0\frac{I_{0}r}{2\mathrm{π}{a}^2}\left(2−\frac{r^2}{a^2}\right)$