Chapter 31, Problem 27PQ

(a)

To determine

Find the expression for magnitudes of the magnetic field in each of the four layers.

Answer to Problem 27PQ

The expression for magnitudes of the magnetic field in each of the four layers are $\underset{\_}{\frac{{\mu }_{0}Ir}{2\pi r_1^2}}$ for $B_1$, $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}}$ for $B_2$, $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}}$ for $B_3$ and $\underset{\_}{0}$ for $B_4$.

Explanation of Solution

Write the expression for Ampere’s Law for the area bounded by the curve as.

    ${\displaystyle \oint B.dl}={\mu }_0{I}_{thru}$

Rearrange above equation for $B$

  $B=\frac{{\mu }_0{I}_{thru}}{{\displaystyle \oint dl}}$                                                                                                   (I)

Here, $B$ is magnetic field, $dl$ is length of element, $I_{thru}$ is current through loop and ${\mu }_0$ is permeability of free space.

Write the expression for current density as.

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

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

Write the expression for area of the circle as.

    $A=\pi R^2$

Substitute $\pi R^2$ for $A$ in equation (II).

    $J=\frac{I}{\pi R^2}$

Consider that the inner coaxial cable of radius $r_1$ and $r$ be the Amperian loop of radius. The current in the loop is product of current density and area of the loop.

Write the expression for Amperian loop current as.

    $I_{thru}=J{A}_1$                                                                                                  (III)

Substitute $\frac{I}{\pi r_1{}^2}$ for $J$ and $\pi r^2$ for $A_1$ in equation (III)

    $I_{thru}=\frac{I{r}^2}{r_1^2}$

Write the expression for Amperian loop is the circumferences of the circle as.

    ${\displaystyle \oint dl}=2\pi r$

Substitute $2\pi r$ for ${\displaystyle \oint dl}$ and $\frac{I{r}^2}{r_1^2}$ for $I_{thru}$ in equation (I).

    $B_1\left(2\pi r\right)={\mu }_0\left(\frac{I{r}^2}{r_1^2}\right)$

Rearrange the above equation as.

    $B_1=\frac{{\mu }_{0}Ir}{2\pi r_1^2}$                                                                                                (IV)

Thus, the magnitude of the magnetic field in inner layer $r_1$ is $\underset{\_}{\frac{{\mu }_{0}Ir}{2\pi r_1^2}}$.

Consider the inside insulator coaxial $r$ be the Amperian loop of radius. In this case, the magnetic field due to inner solenoid is zero. Because the region is outside of the inner solenoid, the magnetic field contribution is due to the outer solenoid.

The current in the loop is product of current density and area of loop is $I_{thru}=I$.

The length of Ameprian loop is equal to the circumferences of the circle that is ${\displaystyle \oint dl}=2\pi r$.

Substitute $2\pi r$ for ${\displaystyle \oint dl}$ and $I$ for $I_{thru}$ in equation (I).

    $B_2=\frac{{\mu }_{0}I}{2\pi r}$                                                                                                   (V)

Thus, the magnitude of the magnetic field in inside insulator layer is $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}}$.

Consider the outer coaxial of radius $r_3$, area of the outer loop is $A_3=\pi \left(r_3^2-r_2^2\right)$ and $r$ be the radius of Amperian loop. Area of the Amperian loop is $A_r=\pi \left(r^2-r_2^2\right)$.

Write the current density of the layer $r_3$ as.

    $J=\frac{I}{A_3}$                                                                                                     (VI)

Substitute $\pi \left(r_3^2-r_2^2\right)$ for $A_3$ in equation (VI)

    $J=\frac{I}{\pi \left(r_3^2-r_2^2\right)}$

Write the expression for Amperian loop current as.

    $I_{thru}=J{A}_r$                                                                                                (VII)

Substitute $\pi \left(r^2-r_2^2\right)$ for $A_r$ and $\frac{I}{\pi \left(r_3^2-r_2^2\right)}$ for $J$ in equation (VII).

    $I_{thru}=\frac{I\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}$

Substitute $2\pi r$ for ${\displaystyle \oint dl}$ and $\frac{I\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}$ for $I_{thru}$ in equation (I).

    $B_3=\frac{{\mu }_{0}I}{2\pi r}\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}$                                                                                (VIII)

Here,$B_3$ is magnetic field in third layer, $I$ is current, $r$ is radius of Gaussian surface, $r_2$ is radius of second layer and $r_3$ is radius for third layer.

Thus, the magnetic field on the third layer is $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}}$.

Write the expression for net magnetic field at the layer $r_3$ is separation between the inside insulator layer field and layer $r_1$ field as.

    $B_{net}=B_2-B_3$                                                                                            (IX)

Substitute $\frac{{\mu }_{0}I}{2\pi r}$ for $B_2$ and $\frac{{\mu }_{0}I}{2\pi r}\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}$ for $B_3$ in equation (IX)

    $B_{net}=\frac{{\mu }_{0}I}{2\pi r}\left[1-\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}\right]$                                                                          (X)

The net current enclosed inside the coaxial cable is zero.

Thus, the magnitude of the magnetic field outside the insulator layer is zero that is $\underset{\_}{B_4=0}$.

Conclusion:

Thus, the expression for magnitudes of the magnetic field in each of the four layers are $\underset{\_}{\frac{{\mu }_{0}Ir}{2\pi r_1^2}}$ for $B_1$, $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}}$ for $B_2$, $\underset{\_}{\frac{{\mu }_{0}I}{2\pi r}\frac{\left(r^2-r_2^2\right)}{\left(r_3^2-r_2^2\right)}}$ for $B_3$ and $\underset{\_}{0}$ for $B_4$.

(b)

To determine

Compare the part a results with the magnetic field produced by a long, straight wire and explain the advantage of using a coax.

Answer to Problem 27PQ

The expression for magnitude of the magnetic field for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.

Explanation of Solution

Write the expression for the magnetic field strength (magnitude) produced by a long straight current-carrying wire as.

    $B=\frac{{\mu }_{0}I}{2\pi r}$

The expression for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.

The advantage of using of coaxial cable as:

1. 1. The inner conductor is in a Faraday shield, noise immunity is improved, and coax has lower error rates and therefore slightly better performance than twisted-pair.
2. 2. Coax provides sufficient frequency range to support multiple channel, which allows for much greater throughput.
3. 3. It also provides greater spacing between amplifiers coax's cable shielding reduces noise and crosstalk

Conclusion:

Thus, the expression for magnitude of the magnetic field for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.