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Chapter 9: Problem 69

A coaxial cable consists of an inner conductor with radius \(r_{i}=0.25 \mathrm{cm}\) and an outer radius of \(r_{\mathrm{o}}=0.5 \mathrm{cm}\) and has a length of 10 meters. Plastic, with a resistivity of \(\rho=2.00 \times 10^{13} \Omega \cdot \mathrm{m},\) separates the two conductors. What is the resistance of the cable?

Short Answer

Expert verified
The resistance of the coaxial cable is approximately \(3.27 \times 10^{17} \Omega\).

Step by step solution

01

Understand the problem

In this problem, we have a coaxial cable which has an inner conductor with radius \(r_i = 0.25 cm\), and an outer conductor with radius \(r_o=0.5 cm\). The cable has a length of 10 meters. The plastic material separating the two conductors has a resistivity of \(\rho = 2.00 \times 10^{13} \Omega m\). We need to find the resistance of the cable.
02

Calculate the resistance of the cable

We know that the formula to find the resistance of a cylindrical conductor is: \[R = \frac{\rho \cdot L}{A}\] Where, \(R\) = resistance \(\rho\) = resistivity of the material \(L\) = length of the wire \(A\) = cross-sectional area of the conductor In this case, we have a coaxial cable. To find the total resistance, we need to find the equivalent resistance for this cable. We know that for coaxial cables, the current is evenly distributed across the conductors. So, the equivalent resistance can be found by considering the resistance of the plastic material for both conductors. Now, let's determine the cross-sectional area for each of the conductors and the resistance of the plastic material for both.
03

Calculate the cross-sectional area for the inner and outer conductors

For cylindrical conductors, the cross-sectional area (A) can be calculated as: \[A_i = \pi{r_i}^2 - \pi{r_i}^2\] \[A_o = \pi{r_o}^2 - \pi{(r_o - r_i)}^2\] Where, \(A_i\) = cross-sectional area of the inner conductor \(A_o\) = cross-sectional area of the outer conductor Now, substitute the given values of \(r_i\) and \(r_o\) to find \(A_i\) and \(A_o\): \[A_i = \pi(0.0025)^2 - \pi(0.0025)^2 = 0\] \[A_o = \pi(0.005)^2 - \pi(0.005 - 0.0025)^2\] \[A_o = \pi(0.005)^2 - \pi(0.0025)^2\approximately 6.11 \times 10^{-5} m^2\]
04

Calculate the resistance of the cable

Now that we have the cross-sectional area of the outer conductor (since the current just flows through the plastic material separating the conductors), let's calculate the resistance of the entire cable using the formula: \[R = \frac{\rho \cdot L}{A_o}\] Substituting the given values of \(\rho\), \(L\), and the calculated value of \(A_o\): \[R = \frac{2.00 \times 10^{13} \Omega m \cdot 10m}{6.11 \times 10^{-5} m^2}\] Now, calculate the resistance: \[R \approx 3.27 \times 10^{17} \Omega\]
05

State the final answer

The resistance of the coaxial cable is approximately \(3.27 \times 10^{17} \Omega\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Conductor Resistance
The resistance in a cylindrical conductor can be understood with the basic formula for resistance: \[ R = \frac{\rho \cdot L}{A} \]Here, \(R\) stands for resistance, \(\rho\) is the resistivity of the material, \(L\) is the length, and \(A\) is the cross-sectional area of the conductor. This formula illustrates that resistance increases with the length of the conductor and decreases as the cross-sectional area increases. In our context of a coaxial cable, resistance plays a crucial role as it directly impacts the efficiency of the cable in signal transmission. Though primarily a conductor, the coaxial cable's resistance is greatly influenced by the insulating material between the conductors. In essence, the overall resistance we calculate includes not just the metal conductors, but also the insulation's resistance to prevent current leakage.
Resistivity
Resistivity is a property of the material that indicates how strongly it opposes the flow of electric current. In our coaxial cable problem, the plastic between the inner and outer conductors has a high resistivity of \[ 2.00 \times 10^{13} \, \Omega\cdot \text{m}\] This shows that the plastic is very resistant to conducting electricity.Materials with high resistivity make excellent insulators and are used in parts of electrical systems where the flow of current needs to be minimized or isolated. For instance, in the coaxial cable, the insulating layer prevents the inner and outer conductors from touching and allows the cable to transmit signals effectively without cross-interference.
Cross-sectional Area Calculation
The cross-sectional area of a cylindrical conductor is critical in understanding how easily current can flow through it. For an ideal cylinder, the area \(A\) can be calculated using the formula: \[ A = \pi r^2 \] Where \(r\) is the radius. In the coaxial cable, however, computing the cross-sectional area involves considering the difference between the two cylindrical surfaces (inner and outer conductors). This is because the plastic insulator fills the space between them, not allowing current to flow directly.For the outer conductor, the formula needs adjustment:\[ A_o = \pi (r_o^2 - (r_o - r_i)^2) \]This equation calculates the annular area, accounting for the conductor and insulation structure. Understanding this concept is key to determining how resistivity and geometry together influence the total resistance of a coaxial cable. Effective use of equations like this helps ensure accurate calculations of cable resistance and efficiency.

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Most popular questions from this chapter

The charge that flows through a point in a wire as a \(\begin{array}{lllll}\text { on } & \text { of } & \text { time } & \text { is } & \text { modeled } & \text { as }\end{array}\) \(q(t)=q_{0} e^{-t / T}=10.0 \mathrm{Ce}^{-t / 5 \mathrm{s}} .\) (a) What is the initial current through the wire at time \(t=0.00 \mathrm{s} ?\) (b) Find the current at time \(t=\frac{1}{2} T .\) (c) At what time \(t\) will the current be reduced by one-half \(I=\frac{1}{2} I_{0} ?\)

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What is the current through a cylindrical wire of radius \(R=0.1 \mathrm{mm}\) if the current density is \(J=\frac{J_{0}}{R} r,\) where \(J_{0}=32000 \frac{\mathrm{A}}{\mathrm{m}^{2}} ?\)
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