b r The diagram below show the inner section of a circular conductor of inner part and outside radius b this transfer current. magnetic field at radial=? a) r< a b) ab

According to Ampere's Law,

$\oint \overrightarrow{B}.\overrightarrow{dl} = {\mu }_0{I}_{enclosed}$

note-

• considering the conductor carries a current I

a) for r<a

The total charged enclosed in the region r<a is zero,

$I_{enclosed }= 0$

using the ampere's law,

$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dl} = {\mu }_0{I}_{enclosed} \\ \oint \overrightarrow{B}.\overrightarrow{dl} = 0 \\ \overrightarrow{B} = 0 \end{gathered}$$

so the magnetic field inside the region r<a is 0.

b) for a<r<b

The current enclosed in a cylindrical shell of inner radius 'a' and outer radius 'r' is 

$$\begin{gathered} I_{enclosed }= \frac{I}{\mathrm{\pi }({\mathrm{b}}^2-{\mathrm{a}}^2)}\times \mathrm{\pi }\left({\mathrm{r}}^2 - {\mathrm{a}}^2\right) \\ {\mathrm{I}}_{\mathrm{enclosed} } = \frac{\mathrm{I}}{({\mathrm{b}}^2-{\mathrm{a}}^2)}\left({\mathrm{r}}^2 - {\mathrm{a}}^2\right) \end{gathered}$$

From ampere's law,

$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dl} = {\mu }_0{I}_{enclosed} \\ \overrightarrow{B}.\oint \overrightarrow{dl} = {\mu }_0\frac{I(r^2-a^2)}{b^2-a^2} \\ \overrightarrow{B}\left(2\mathrm{\pi r}\right) = {\mu }_0\frac{I(r^2-a^2)}{b^2-a^2} \\ \overrightarrow{B} =\frac{ {\mu }_0}{2\mathrm{\pi r}}\frac{I(r^2-a^2)}{b^2-a^2} \end{gathered}$$