91Ó°ÊÓ

• The radius of the central solid cylindrical part of the cable$c=0.40\text{cm}=0.40×{10}^{-2}\text{m}$ .
• The inner radius of the outer hollow cylindrical part of the cable $b=1.8\text{cm}=1.8×{10}^{-2}\text{m}$.
• The outer radius of the outer hollow cylindrical part of the cable$a=2\text{cm}=2.0×{10}^{-2}\text{m}$ .
• Current through inner and outer cylindrical part i = 120 A flowing in the opposite sense.

For this problem, we can use Ampere’s law by considering a suitable Ampere loop to find the magnetic field at the given point over the specified region. After that, we can find out the current enclosed in the ampere loop. Finally, we get the expression for the Bfield at the particular point.

Formula:

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

For r < c,

$\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}& =& {\mathrm{μ}}_0{I}_{enclosed}\\ Bâˆ\circledR ds& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B·2\mathrm{Ï€}r& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B& =& \frac{{\mathrm{μ}}_0{I}_{enclosed}}{2\mathrm{Ï€}r}\\ & & \end{array}$

Now let’s findthecurrent enclosed in the ampere loop of radius r:

$\begin{array}{rcl}{I}_{enclosed}& =& \frac{\mathrm{Ï€}{r}^2}{\mathrm{Ï€}{c}^2}i\\ & =& \frac{r^2}{c^2}i\\ & & \end{array}$

So we get,

$B=\frac{{\mathrm{μ}}_0}{2\mathrm{π}r}\frac{r^2}{c^2}=\frac{{\mathrm{μ}}_{0}ir}{2\mathrm{π}{c}^2}$

Hence,$B=\frac{{\mathrm{μ}}_{0}ir}{2\mathrm{π}{c}^2}.$

Forc < r < b

Now let’s find the current enclosed in the ampere loop of radius r:

$\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}& =& {\mathrm{μ}}_0{I}_{enclosed}\\ Bâˆ\circledR ds& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B·2\mathrm{Ï€}r& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B& =& \frac{{\mathrm{μ}}_0{I}_{enclosed}}{2\mathrm{Ï€}r}\\ & & \end{array}$

$I_{enclosed}=i$, so we get$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

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

Forb < r < a

$\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}& =& {\mathrm{μ}}_0{I}_{enclosed}\\ Bâˆ\circledR ds& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B·2\mathrm{Ï€}r& =& {\mathrm{μ}}_0{I}_{enclosed}\\ B& =& \frac{{\mathrm{μ}}_0{I}_{enclosed}}{2\mathrm{Ï€}r}\\ & & \end{array}$

Let’s findthecurrent enclosed in the ampere loop of radiusr:

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

Hence using this, we get $B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\frac{a^2-r^2}{a^2-b^2}$.

For r > a

Current through the inner and outer cylindrical part is i flowing in the opposite direction, so the current enclosed in the Ampere loop of radius r

$I_{enclosed}=0.0$

Hence $B=0.0$.

The B (r)for the special cases:

We have

$B\left(r\right)=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\frac{a^2-r^2}{a^2-b^2}$

When a = 0, we get

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

Whenr = b,we get

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

When r = a, we get

$\begin{array}{rcl}B\left(r\right)& =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}a}·\frac{a^2-a^2}{a^2-b^2}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}a}·0\\ & =& 0.0\\ & & \end{array}$

When b = 0, we get

$\begin{array}{rcl}B\left(r\right)& =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\frac{a^2-r^2}{a^2-0}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}·\left(1-\frac{r^2}{a^2}\right)\\ & & \end{array}$

Hence, it is tested the expressions for all special cases.

The plot of the function B (r) over the range 0 < r < 3 cm and $i=120\text{A:}$$c=0.40\text{cm}=0.40×{10}^{-2}\text{m}$, $b=1.8\text{cm}=1.8×{10}^{-2}\text{m}$,$a=2\text{cm}=2.0×{10}^{-2}\text{m}$.

Hence the graph is drawn.