Chapter 19, Problem 60E

(a)

To determine

The magnetic field for a current carrying hollow tube in the region $r<b$.

Answer to Problem 60E

The magnetic field in the region $r<b$ is zero.

Explanation of Solution

Write the expression for the current density due to hollow tube.

    $J=\frac{I}{A}$        (1)

Here, $J$ is the current density, $I$ is the current and $A$ is the area. $a$ is the inner radius of the tube and $b$ is the outer radius of the tube.

Write the expression Ampere’s law.

    ${\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_{0}I$        (2)

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

Conclusion:

Substitute $0$ for $I$ in equation (2).

    $\begin{array}{c}{\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}=0\\ B=0\end{array}$

Thus, the magnetic field in the region $r<b$ is zero.

(b)

To determine

The magnetic field for a current carrying hollow tube in the region $b<r<c$.

Answer to Problem 60E

The magnetic field in the region $b<r<c$ is $\frac{{\mu }_{0}I\left(r^2-b^2\right)}{2\pi \left(c^2-b^2\right)r}$.

Explanation of Solution

Write the expression for the current density due to hollow tube.

    $J=\frac{I}{A}$

Write the expression Ampere’s law.

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

Conclusion:

The current density for the tube is calculated below.

Substitute $\pi \left(c^2-b^2\right)$ for $A$ in equation (1).

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

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

Total current in the region $b<r<c$ is calculated below.

Substitute $I^{\prime }$ for $I$$\frac{I}{\pi \left(c^2-b^2\right)}$ for $J$ and $\pi \left(r^2-b^2\right)$ for $A$ in equation (1) and simplify.

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

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

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

    $\begin{array}{c}{\displaystyle \oint \overrightarrow{B}}\cdot \overrightarrow{dl}={\mu }_0\frac{I\pi \left(r^2-b^2\right)}{\pi \left(c^2-b^2\right)}\\ B{\displaystyle \oint dl}={\mu }_0\frac{I\pi \left(r^2-b^2\right)}{\pi \left(c^2-b^2\right)}\end{array}$

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

    $\begin{array}{c}B\left(2\pi r\right)={\mu }_0\frac{I\pi \left(r^2-b^2\right)}{\pi \left(c^2-b^2\right)}\\ B=\frac{{\mu }_{0}I\left(r^2-b^2\right)}{2\pi \left(c^2-b^2\right)r}\end{array}$

Thus, the magnetic field in the region $b<r<c$ is $\frac{{\mu }_{0}I\left(r^2-b^2\right)}{2\pi \left(c^2-b^2\right)r}$.

(c)

To determine

The magnetic field in the region $r>c$ is $\frac{{\mu }_{0}I}{2\pi r}$.

Explanation of Solution

Write the expression for the current density due to hollow tube.

    $J=\frac{I}{A}$

Write the expression Ampere’s law.

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

Conclusion:

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

    $B\left(2\pi r\right)={\mu }_{0}I$

Simplify the above equation.

    $B=\frac{{\mu }_{0}I}{2\pi r}$

Thus, the magnetic field in the region $r>c$ is $\frac{{\mu }_{0}I}{2\pi r}$.