Chapter 19, Problem 59E

(a)

To determine

The magnetic field for two conductors in region $r<a$.

Answer to Problem 59E

The magnetic field in region $r<a$ is $\frac{2k^{\prime }Ir}{a^2}$.

Explanation of Solution

Write the expression for Ampere’s law.

    ${\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$        (1)

Here, $B$ is the magnetic field, $dl$ is an elemental length, ${\mu }_0$ is the free space permeability and $I$ is the current.

Write the expression for current density.

    $J=\frac{I}{A}$        (2)

Here, $J$ is the current density and $A$ is the surface area.

Conclusion:

In the region $r<a$ coaxial cable is only current contributor

Substitute $\left(\pi a^2\right)$ for $A$ in equation (2).

    $J=\frac{I}{\pi a^2}$

Here, $a$ is the radius of the cable.

Substitute $I^{\prime }$ for $I$ and $\left(\pi r^2\right)$ for $A$ in equation 2).

    $J=\frac{I^{\prime }}{\pi r^2}$

Here, $I^{\prime }$ is the current in region $r<a$ and $r$ is the distance of a point in that region.

Substitute $\frac{I}{\pi a^2}$ for $J$ in the above equation.

    $\frac{I}{\pi a^2}=\frac{I^{\prime }}{\pi r^2}$

Rearrange the above equation.

    $\begin{array}{c}{I}^{\prime }=\frac{I\pi r^2}{\pi a^2}\\ =\frac{I{r}^2}{a^2}\end{array}$

Substitute $\frac{I{r}^2}{a^2}$ for $I$ in equation (1).

    $B{\displaystyle \oint dl}={\mu }_0\frac{I{r}^2}{a^2}$

Substitute $\left(2\pi r\right)$ for the close integral in the above equation.

    $B\left(2\pi r\right)=\frac{{\mu }_0}{4\pi }\frac{\left(4\pi \right)I{r}^2}{a^2}$

Substitute $k^{\prime }$ for $\frac{{\mu }_0}{4\pi }$ in the above expression.

    $B=\frac{2k^{\prime }Ir}{a^2}$

Thus, the magnetic field in region $r<a$ is $\frac{2k^{\prime }Ir}{a^2}$.

(b)

To determine

The magnetic field for two conductors in region $a<r<b$.

Answer to Problem 59E

The magnetic field in region $a<r<b$ is $\frac{2k^{\prime }I}{r}$.

Explanation of Solution

Write the expression for Ampere’s law.

    ${\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$

Conclusion:

Rearrange equation (1).

    $B{\displaystyle \oint dl}={\mu }_{0}I$

Substitute $\left(2\pi r\right)$ for the close integral and $4\pi k^{\prime }$ for ${\mu }_0$ in the above equation.

    $\begin{array}{c}B\left(2\pi r\right)=\left(4\pi k^{\prime }\right)I\\ B=\frac{2k^{\prime }I}{r}\end{array}$

Here, $k^{\prime }$ is a constant and $r$ is the distance of a point in that region.

Thus, the magnetic field in region $a<r<b$ is $\frac{2k^{\prime }I}{r}$.

(c)

To determine

The magnetic field in region $b<r<c$ is $\frac{2k^{\prime }I\left(r^2-b^2\right)}{r\left(c^2-b^2\right)}$.

Explanation of Solution

Write the expression for Ampere’s law.

    ${\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$

Write the expression for current density.

    $J=\frac{I}{A}$

Conclusion:

In the region $b<r<c$ tube is only current contributor

Substitute $\pi \left(c^2-b^2\right)$ for $A$ in equation (2).

    $J=\frac{I}{\pi \left(c^2-b^2\right)}$

Here, $b$ is the inner radius of the tube and $c$ is the outer radius of the tube.

Substitute $I^{\prime }$ for $I$ and $\pi \left(r^2-b^2\right)$ for $A$ in equation 2).

    $J=\frac{I^{\prime }}{\pi \left(r^2-b^2\right)}$

Here, $I^{\prime }$ is the current in region $b<r<c$ and $r$ is the distance of a point in that region.

Substitute $\frac{I}{\pi \left(c^2-b^2\right)}$ for $J$ in the above equation.

    $\frac{I}{\pi \left(c^2-b^2\right)}=\frac{I^{\prime }}{\pi \left(r^2-b^2\right)}$

Rearrange the above equation.

    $\begin{array}{c}{I}^{\prime }=\frac{I\pi \left(r^2-b^2\right)}{\pi \left(c^2-b^2\right)}\\ =\frac{I\left(r^2-b^2\right)}{\left(c^2-b^2\right)}\end{array}$

Substitute $\frac{I\left(r^2-b^2\right)}{\left(c^2-b^2\right)}$ for $I$ in equation (1).

    $B{\displaystyle \oint dl}={\mu }_0\frac{I\left(r^2-b^2\right)}{\left(c^2-b^2\right)}$

Substitute $\left(2\pi r\right)$ for the close integral in the above equation.

    $B\left(2\pi r\right)=\frac{{\mu }_0}{4\pi }\frac{\left(4\pi \right)I\left(r^2-b^2\right)}{\left(c^2-b^2\right)}$

Substitute $k^{\prime }$ for $\frac{{\mu }_0}{4\pi }$ and solve the above expression.

    $B=\frac{2k^{\prime }I\left(r^2-b^2\right)}{r\left(c^2-b^2\right)}$

Thus, the magnetic field in region $b<r<c$ is $\frac{2k^{\prime }I\left(r^2-b^2\right)}{r\left(c^2-b^2\right)}$.

(d)

To determine

The magnetic field for two conductors in region $r>c$.

Answer to Problem 59E

The magnetic field in region $r>c$ is $0$.

Explanation of Solution

Write the expression for Ampere’s law.

    ${\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$

Conclusion:

In the region $r>c$ there is no contribution for current.

Substitute $0$ for the $I$ in equation (1).

    $\begin{array}{c}{\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_0\left(0\right)\\ B=0\end{array}$

Thus, the magnetic field in region $r>c$ is $0$.