Question

A long, hollow wire has inner radius $R_1$and outer radius localid="1649505313415" $R_2$. The wire carries current I uniformly distributed across the area of the wire. Use Ampère’s law to find an expression for the magnetic field strength in the three regions $0\le r\le R_1,R_1\le r\le R_{2}and{R}_2\le r$.

Short answer

$$\begin{gathered} B_{0\le r\le R_1}=0 \\ {B}_{R1\le r\le R_2}=\frac{{\mu }_{o}I}{2\pi r}\left(\frac{r^2-R^2}{R_2^2-R_1^2}\right), \\ {B}_{r\ge R_2}=\frac{{\mu }_{o}I}{2\pi r} \end{gathered}$$

Step-by-step solution

Given Information

By using Ampère’s law, we are finding expression for the magnetic field strength in the three regions.

Simplify

The circumference of the enclosed area to get the magnetic field of Ampère’s law.

$$\begin{gathered} {\int }_0^l\overrightarrow{B}·dl={\mu }_o{I}_{encl} \\ B{\left[l\right]}_0^{2\pi r}={\mu }_o{I}_{encl} \\ B=\frac{{\mu }_o{I}_{encl}}{2\pi r} \end{gathered}$$

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

$\boxed{B_{0\le r\le R_1}=0}$

Simplification.

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

$R_1\le r\le R_2$

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

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

$J_r=\frac{I_{encl}}{\pi r^2}$

We will get the current $I_{encl}$

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

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

$B_{R_1\le r\le R_2}=\frac{{\mu }_{o{I}_{encl}}}{2\pi r}=\frac{{\mu }_{o\left({\displaystyle \frac{r^2-R^2}{R_2^2-R_1^2}}I\right)}}{2\pi r}=\boxed{\frac{{\mu }_{o}I}{2\pi r}\left(\frac{r^2-R^2}{R_2^2-R_1^2}\right)}$

The distance $r\ge R_2$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.

$$\begin{gathered} B{\left[l\right]}_0^{2\pi r}={\mu }_{o}I \\ B\left(2\pi r\right)={\mu }_{o}I \\ \boxed{B_{r\ge R_2}=\frac{{\mu }_{o}I}{2\pi r}} \end{gathered}$$