Chapter 30, Problem 75CP

(a)

To determine

The magnitude and direction of magnetic field at point $P_1$.

Answer to Problem 75CP

The magnetic field at point $P_1$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ directed to the left.

Explanation of Solution

Write the expression for the current density.

  $J=\frac{I}{A}$                                                                                                                  (I)

Here, $J$ is the current density, $I$ is the current in the conductor and $A$ is the area of conductor.

Write the expression for the area of the conductor.

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

Substitute $\frac{\pi a^2}{2}$ for $A$ in equation (I) to find $J$.

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

Write the expression for the magnetic field in the solid portion.

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

Here, $B_{\text{solid}}$ is the magnetic field in the solid portion and ${\mu }_0$ is the permeability of the free space.

Write the expression for the magnetic field at point 1.

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

Write the expression for the magnetic field at point 2.

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

Write the expression of net magnetic field at point $P_1$.

  $B=B_{\text{solid}}-B_1-B_2$                                                                                            (VII)

Here, $B$ is the net magnetic field at point $P_1$.

Substitute $\frac{{\mu }_{0}J\left(\pi a^2\right)}{2\pi r}$ for $B_{\text{solid}}$, $\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \left(r-\left(a/2\right)\right)}$ for $B_{\text{1}}$ and $\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \left(r+\left(a/2\right)\right)}$ for $B_{\text{2}}$ in equation (VII) to find $B$.

    $B=\frac{{\mu }_{0}J\left(\pi a^2\right)}{2\pi r}-\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \left(r-\left(a/2\right)\right)}-\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \left(r+\left(a/2\right)\right)}$                                       (VIII)

Substitute $\frac{2I}{\pi a^2}$ for $J$ in equation (VIII) to find $B$.

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

Conclusion:

Therefore, the magnetic field at point $P_1$ is $\frac{{\mu }_{0}I}{\pi r}\left(\frac{2r^2-a^2}{4r^2-a^2}\right)$ directed to the left.

(b)

To determine

The magnitude and direction of magnetic field at point $P_2$.

Answer to Problem 75CP

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

Explanation of Solution

Write the expression for the magnetic field at point 1 and point 2.

  $\begin{array}{c}{B}_1=\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \sqrt{r^2+{\left(a/2\right)}^2}}\\ =B_2\end{array}$                                                                                      (IX)

Write the expression for the net magnetic field at point $P_2$.

  $B=B_{\text{solid}}-B_1\cos \theta -B_2\cos \theta$                                                                              (X)

Substitute $\frac{{\mu }_{0}J\left(\pi a^2\right)}{2\pi r}$ for $B_{\text{solid}}$, $\frac{{\mu }_{0}J\pi {\left(a/2\right)}^2}{2\pi \sqrt{r^2+{\left(a/2\right)}^2}}$ for $B_{\text{1}}$ and $B_{\text{2}}$, and $\left(\frac{r}{\sqrt{r^2+\left(a^2/4\right)}}\right)$ for $\cos \theta$ in equation (X) to find $B$.

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

Conclusion:

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