Question

Figure 29-87 shows a cross section of a hollow cylindrical conductor of radii aand b, carrying a uniformly distributed currenti. (a) Show that the magnetic field magnitude B(r) for the radial distancer in the range$\mathit{b}\mathbf{}\mathbf{<}\mathbf{}\mathit{r}\mathbf{}\mathbf{<}\mathbf{}\mathit{a}$is given by$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi }\mathbf{r}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}\mathbf{·}\frac{(r^2-b^2)}{\mathbf{r}}\mathbf{}$

(b) Show that when r = a, this equation gives the magnetic field magnitude Bat the surface of a long straight wire carrying current i; when r = b, it gives zero magnetic field; and when b = 0, it gives the magnetic field inside a solid conductor of radius acarrying current i. (c) Assume that a = 2.0 cm, b = 1,8 cm, and i = 100 A, and then plot B(r) for the range $\mathbf{0}\mathbf{}\mathbf{<}\mathbf{}\mathit{r}\mathbf{}\mathbf{<}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ .

Short answer

All the equations have been proved.

Step-by-step solution

Given data

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.

Understanding the concept

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:

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

(a) Show that the magnitude of the magnetic field at radial distance r  is B=μ0i2πr·( r2-b2 )a2-b2  

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

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

$B\oint ds=B2\pi r={\mu }_0{I}_{enclosed}$

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

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

(b) Show that When r = a this equation gives the magnetic field magnitude B at the surface of a long straight wire carrying a current i , and when b = 0, it gives the magnetic field inside a solid conductor of radius a carrying current i.

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

$$\begin{gathered} B=\frac{{\mu }_{0}i\left(\frac{a^2-b^2}{a^2-b^2}\right)}{2\pi a} \\ B=\frac{{\mu }_{0}i}{2\pi a}\dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right) \end{gathered}$$

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

$B=\frac{{\mu }_{0}i}{2\pi r}·\frac{r^2}{a^2}=\frac{{\mu }_{0}ir}{2\pi a^2}$

(c) Plot B(r) for the range  0 < r < 6.0 cm

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

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

$B=\frac{{\mu }_{0}i}{2\pi r}$

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