91Ó°ÊÓ

1. The inner radius of the hollow conducting cylinder is b.
2. The outer radius of the hollow conducting cylinder is a.
3. The radial distance from the center of the hollow conducting cylinder is r.
4. The current flowing through the hollow conducting cylinder is i = 100 A.

We can use Ampere’s law to prove the statement given in the problem. For this, we must find out the current enclosed in a specific region. Therefore, in the Ampere loop also, we will consider another Ampere loop through the point where we want to find the field.

Formula:

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0{I}_{enclosed}$

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0{I}_{enclosed}$

Let r be the point at which we want to find out the field.

$Bâˆ\circledR ds=B2\mathrm{Ï€}r={\mathrm{μ}}_0{I}_{enclosed}$

Now we will determine the current enclosed in the ampere loop. Refer figure 27-87. We get

$\begin{array}{rcl}{I}_{enclosed}& =& \frac{\mathrm{π}{r}^2-\mathrm{π}{b}^2}{\mathrm{π}{a}^2-\mathrm{π}{b}^2}·i=\frac{r^2-b^2}{a^2-b^2}·i\\ B2\mathrm{π}r& =& {\mathrm{μ}}_0\frac{r^2-b^2}{a^2-b^2}·i\\ B& =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\left(\frac{r^2-b^2}{a^2-b^2}\right)\\ & & \end{array}$

When r = a using the above expression, we get the magnitude of magnetic field at the surface of the long straight wire

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}i\left(\frac{a^2-b^2}{a^2-b^2}\right)}{2\mathrm{π}a} \\ B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}a}…………………….\left(1\right) \end{gathered}$$

When r = bwe getthe magnetic field at the inner surface of a hollow cylindrical conductor asB = 0.

When b = 0, we get the magnitude of the magnetic field inside a solid conductor of radius a

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\frac{r^2}{a^2}=\frac{{\mathrm{μ}}_{0}ir}{2\mathrm{π}{a}^2}$

PlotB (r)for the range0 < r < 6.0 cm

The field inside the conductor is zero forr < band can be found from

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

Using this and the result found in part a), we can plot the graph.