Chapter 30, Problem 30.75CP

(a)

To determine

The magnitude and direction of the magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_1$.

Answer to Problem 30.75CP

The magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_1$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ and directed towards the left side.

Explanation of Solution

Given info: The given figure is shown below:

Figure (1)

Write the expression for the current in the solid conductor is,

$I=JA$

Here,

$J$ is the current density.

$I$ is the current flows in the conductor.

$A$ is the area of the conducting part.

Formula to calculate the net area of the conductor is,

$\begin{array}{c}A=\pi \left(a^2-{\left(\frac{a}{2}\right)}^2-{\left(\frac{a}{2}\right)}^2\right)\\ =\frac{\pi a^2}{2}\end{array}$

Replace $A$ by $\frac{\pi a^2}{2}$ in equation (1).

$I=\frac{J\pi a^2}{2}$

Rearrange for $J$.

$J=\frac{2I}{\pi a^2}$

Write the expression for the area of the solid conducting part is,

$A_{\text{s}}=\pi a^2$

Here,

$a$ is the radius of the conducting cylinder.

Write the expression for the area of the cavity part is,

$\begin{array}{c}{A}_{\text{c1}}=\pi {\left(\frac{a}{2}\right)}^2\\ =\frac{\pi a^2}{4}\end{array}$

From the right hand rule, the current is directed out of the page so, the direction of magnetic field at point $P_1$ is towards the left side.

Formula to calculate the magnetic field produces at point $P_1$ considering the conductor as a whole solid cylinder is,

$B_{\text{s}}=\frac{{\mu }_{0}J{A}_{\text{s}}}{2\pi r}$

Here,

${\mu }_0$ is the absolute permeability.

Substitute $\pi a^2$ for $A_{\text{s}}$ in above equation.

$\begin{array}{c}{B}_{\text{s}}=\frac{{\mu }_{0}J\left(\pi a^2\right)}{2\pi r}\\ =\frac{{\mu }_{0}J{a}^2}{2r}\end{array}$

Formula to calculate the magnetic field produces at point $P_1$ due to cavity is,

$B_{\text{1}}=\frac{{\mu }_{0}J{A}_{\text{c1}}}{2\pi \left(r-\frac{a}{2}\right)}$

Substitute $\frac{\pi a^2}{4}$ for $A_{\text{c1}}$ in above equation.

$\begin{array}{c}{B}_{\text{1}}=\frac{{\mu }_{0}J\left(\frac{\pi a^2}{4}\right)}{2\pi \left(r-\frac{a}{2}\right)}\\ =\frac{{\mu }_0{a}^2}{8\left(r-\frac{a}{2}\right)}\end{array}$

Since both the cavities are of same hence the field will be same.

$B_{\text{2}}=\frac{{\mu }_{0}J{a}^2}{8\left(r+\frac{a}{2}\right)}$

Formula to calculate the net magnetic field produces at point $P_1$ is,

$B=B_{\text{s}}-B_1-B_2$

Substitute $\frac{{\mu }_{0}J{a}^2}{2r}$ for $B_{\text{s}}$, $\frac{{\mu }_{0}J{a}^2}{8\left(r-\frac{a}{2}\right)}$ for $B_{\text{1}}$ and $\frac{{\mu }_{0}J{a}^2}{8\left(r+\frac{a}{2}\right)}$ for $B_{\text{2}}$.

$\begin{array}{c}B=\frac{{\mu }_{0}J{a}^2}{2r}-\frac{{\mu }_{0}J{a}^2}{8\left(r-\frac{a}{2}\right)}-\frac{{\mu }_{0}J{a}^2}{8\left(r+\frac{a}{2}\right)}\\ =\frac{{\mu }_{0}J{a}^2}{2}\left(\frac{1}{r}-\frac{1}{4\left(r-\frac{a}{2}\right)}-\frac{1}{4\left(r+\frac{a}{2}\right)}\right)\\ =\frac{{\mu }_{0}J{a}^2}{2}\left(\frac{4r^2-a^2-2r^2}{4r\left(r^2-a^2/4\right)}\right)\\ =\frac{{\mu }_{0}J{a}^2}{2\pi }\left(\frac{2r^2-a^2}{4r\left(r^2-a^2/4\right)}\right)\end{array}$

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

$\begin{array}{c}B=\frac{{\mu }_0\left(\frac{2I}{\pi a^2}\right)a^2}{2}\left(\frac{2r^2-a^2}{4r\left(r^2-a^2/4\right)}\right)\\ =\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)\end{array}$

Conclusion:

Therefore, the magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_1$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ and directed towards the left side.

(b)

To determine

The magnitude and direction of the magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_2$.

Answer to Problem 30.75CP

The magnitude of the magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_2$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2+a^2}{4r^2+a^2}\right)$ and directed towards the top of the page.

Explanation of Solution

From the right hand rule, the current is directed out of the page so, the direction of magnetic field at point $P_2$ is towards the top of the page.

Formula to calculate the magnetic field produces at point $P_2$ considering the conductor as a whole solid cylinder is,

$B_{\text{s}}=\frac{{\mu }_{0}J{A}_{\text{s}}}{2\pi r}$

Here,

${\mu }_0$ is the absolute permeability.

Substitute $\pi a^2$ for $A_{\text{s}}$ in above equation.

$\begin{array}{c}{B}_{\text{s}}=\frac{{\mu }_{0}J\left(\pi a^2\right)}{2\pi r}\\ =\frac{{\mu }_{0}J{a}^2}{2r}\end{array}$

Formula to calculate the magnetic field produces at point $P_2$ due to cavity is,

$B_{\text{1}}=\frac{{\mu }_{0}J{A}_{\text{c1}}}{2\pi \sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}$

Substitute $\frac{\pi a^2}{4}$ for $A_{\text{c1}}$ in above equation.

$\begin{array}{c}{B}_{\text{1}}=\frac{{\mu }_{0}J\left(\frac{\pi a^2}{4}\right)}{2\pi \sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}\\ =\frac{{\mu }_{0}J{a}^2}{8\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}\end{array}$

Since both the cavities are of same hence the field will be same.

$B_{\text{2}}=\frac{{\mu }_{0}J{a}^2}{8\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}$

Formula to calculate the net magnetic field produces at point $P_1$ is,

$B=B_{\text{s}}-B_1\cos \theta -B_2\cos \theta$

Substitute $\frac{{\mu }_{0}J{a}^2}{2r}$ for $B_{\text{s}}$, $\frac{r}{\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}$ for $\cos \theta$, $\frac{{\mu }_{0}J{a}^2}{8\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}$ for $B_{\text{1}}$ and for $B_{\text{2}}$.

$\begin{array}{c}B=\frac{{\mu }_{0}J{a}^2}{2r}-\left(\frac{{\mu }_{0}J{a}^2}{8\left(r-\frac{a}{2}\right)}\right)\left(\frac{r}{\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}\right)-\left(\frac{{\mu }_{0}J{a}^2}{8\left(r+\frac{a}{2}\right)}\right)\left(\frac{r}{\sqrt{r^2+{\left(\frac{a}{2}\right)}^2}}\right)\\ =\frac{{\mu }_{0}J{a}^2}{2r}\left(1-\frac{r^2}{2\left(r^2+a^2/4\right)}\right)\end{array}$

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

$\begin{array}{c}B=\frac{{\mu }_0\left(\frac{2I}{\pi a^2}\right)a^2}{2r}\left(1-\frac{r^2}{2\left(r^2+a^2/4\right)}\right)\\ =\frac{{\mu }_{0}I}{\pi r}\left(1-\frac{r^2}{2\left(r^2+a^2/4\right)}\right)\\ =\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2+a^2}{4r^2+a^2}\right)\end{array}$

Conclusion:

Therefore, the magnitude of the magnetic field in terms of ${\mu }_0$, $I$, $r$ and $a$ at a point $P_2$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2+a^2}{4r^2+a^2}\right)$ and directed towards the top of the page.