91影视

By using Amp猫re鈥檚 law, we are finding expression for the magnetic field strength in the three regions.

The circumference of the enclosed area to get the magnetic field of Amp猫re鈥檚 law.

$$\begin{gathered} {鈭�}_0^l\stackrel{鈫�}{B}路dl={渭}_o{I}_{encl} \\ B{\left[l\right]}_0^{2蟺r}={渭}_o{I}_{encl} \\ B=\frac{{渭}_o{I}_{encl}}{2蟺r} \end{gathered}$$

Point $0鈮�r鈮�R_1$the enclosed current is zero, So, the magnetic field in this region is 0.

$\boxed{B_{0鈮�r鈮�R_1}=0}$

The current density inside the enclosed area (r) equals the current density in the whole wire of radius $R_1$.

$R_1鈮�r鈮�R_2$

$J_1=\frac{I}{蟺\left(R_2^2-R_1^2\right)}$

$I$is the current and the current density for the radial path.

$J_r=\frac{I_{encl}}{蟺r^2}$

We will get the current $I_{encl}$

$$\begin{gathered} J_1=J_r \\ \frac{I}{蟺\left(R_2^2-R_1^2\right)}=\frac{I_{encl}}{蟺\left(r^2-R^2\right)} \\ {I}_{encl}=\frac{r^2-R^2}{\left(R_2^2-R_1^2\right)}I \end{gathered}$$

The expression for $I$into equation (1) $B_{R_1鈮�r鈮�R_2}$

$B_{R_1鈮�r鈮�R_2}=\frac{{渭}_{o{I}_{encl}}}{2蟺r}=\frac{{渭}_{o\left({\displaystyle \frac{r^2-R^2}{R_2^2-R_1^2}}I\right)}}{2蟺r}=\boxed{\frac{{渭}_{o}I}{2蟺r}\left(\frac{r^2-R^2}{R_2^2-R_1^2}\right)}$

The distance $r鈮�R_2$is the same for the current of the wire with the radius $R_1$, so let us integrate over the circumference of the enclosed area to get the magnetic field.

$$\begin{gathered} B{\left[l\right]}_0^{2蟺r}={渭}_{o}I \\ B\left(2蟺r\right)={渭}_{o}I \\ \boxed{B_{r鈮�R_2}=\frac{{渭}_{o}I}{2蟺r}} \end{gathered}$$