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Chapter 28: Problem 45

Coaxial Cable. A solid conductor with radius \(a\) is supported by insulating disks on the axis of a conducting tube with inner radius \(b\) and outer radius \(c\) (Fig. \(\mathrm{E} 28.45\) ). The central conductor and tube carry equal currents \(I\) in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube \((ac) .\)

Short Answer

Expert verified
(a) \( \mathbf{B} = \frac{\mu_0 I}{2 \pi r} \) for \( a < r < b \); (b) \( \mathbf{B} = 0 \) for \( r > c \).

Step by step solution

01

Set Up Ampere's Law for Region a < r < b

To derive the magnetic field at a distance r, where a < r < b, we'll use Ampere's Law:\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \]Here, \(I_{enc}\) is the net current enclosed by the Amperian loop, a circle of radius r around the wire. In this region, the enclosed current is the full current I through the central conductor. Thus:\[ \mathbf{B}(2 \pi r) = \mu_0 I \]Solving for \(\mathbf{B}\), we get:\[ \mathbf{B} = \frac{\mu_0 I}{2 \pi r} \]
02

Analyze Magnetic Field for Region r > c

For this region, we still apply Ampere's Law, but at \(r > c\), the currents on the central conductor and tube cancel out since each carries an equal but opposite current \(I\). Therefore, the net current enclosed is zero:\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I - I) = 0 \]So, the magnetic field is:\[ \mathbf{B} = 0 \]

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Key Concepts

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

Magnetic Field
The magnetic field is essential in understanding the electromagnetic force around objects like wires and cables. It represents the force that a magnetic object experiences in the presence of another magnetic object or current. When dealing with current-carrying wires, Ampere's Law helps us calculate the magnetic field that forms around them.

Using Ampere's Law, we can determine how the magnetic field behaves within different regions, such as inside or outside certain boundaries. In the context of our coaxial cable problem, the law helps us derive the magnetic field in specific regions around the cable. The field inside a circular path with radius between the inner wire and outer tube \(a < r < b\) is calculated as \(B = \frac{\mu_0 I}{2 \pi r}\)\. On the other hand, outside the entire structure \(r > c\), the magnetic field \(B\) is zero, since the opposing currents cancel out completely.

Key points to understand about magnetic fields in this setup include:
  • They form concentric circles around the current path.
  • The strength decreases with distance from the wire.
  • Internal cancellation in coaxial designs can negate external fields.
Coaxial Cable
A coaxial cable is a type of electrical cable that consists of a central conductor surrounded by an insulating spacer, enclosed in a cylindrical conducting shield. The entire cable is insulated on the outside.

This construction ensures that the electromagnetic fields generated by the currents within the cable remain confined, reducing interference from external fields and minimizing the emission of signals outside the cable. This quality is essential in electronics, especially for transmitting signals with high fidelity, such as in cable television and internet data.

The coaxial cable setup in our exercise involves the flow of equal and opposite currents through the central conductor and the outer tube. This arrangement causes important effects:
  • Reduction of external magnetic fields due to cancellation.
  • Uniform distribution of current over cross sections.
  • Ability to carry a signal over long distances with minimal loss.
Current Distribution
Current distribution refers to how the electric current is spread across the cross-sectional area of a conductor. In the context of the exercise, the central conductor and the outer tube of the coaxial cable both carry current uniformly.

This uniformity means that the current density remains consistent throughout the material. Current density is the amount of current flowing per unit area, and in our case, it is evenly spread in both the inner and outer conductors, ensuring smooth power or data transmission.

Understanding the implications of uniform current distribution helps in recognizing:
  • How the magnetic field is calculated using Ampere's Law.
  • The nullifying effect of opposing currents in coaxial cables.
  • The influence of uniform current on minimizing losses and enhancing efficiency.
Practical designs utilizing uniform current distribution prevent overheating and maximize reliability, making coaxial cables effective for various communication technologies.

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Most popular questions from this chapter

A toroidal solenoid has an inner radius of 12.0 \(\mathrm{cm}\) and an outer radius of 15.0 \(\mathrm{cm} .\) It carries a current of 1.50 A. How many equally spaced turns must it have so that it will produce a magnetic field of 3.75 mT at points within the coils 14.0 \(\mathrm{cm}\) from its center?

Lamp Cord Wires. The wires in a household lamp cord are typically 3.0 \(\mathrm{mm}\) apart center to center and carry equal currents in opposite directions. If the cord carries current to a \(100-\mathrm{W}\) light bulb connected across a \(120-\mathrm{V}\) potential difference, what force per meter does each wire of the cord exert on the other? Is the force attractive or repulsive? Is this force large enough so it should be considered in the design of the lamp cord? (Model the lamp cord as a very long straight wire.)

BIO Currents in the Heart. The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field around the chest due to currents in the heart give values of about 10\(\mu \mathrm{G}\) . Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire. If the surface of the chest is 5.0 \(\mathrm{cm}\) from this current, how large is the current in the heart?

At a particular instant, charge \(q_{1}=+4.80 \times 10^{-6} \mathrm{C}\) is at the point \((0,0.250 \mathrm{m}, 0)\) and has velocity \(\vec{\boldsymbol{v}}_{1}=\left(9.20 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\imath}\) Charge \(q_{2}=-2.90 \times 10^{-6} \mathrm{C}\) is at the point \((0.150 \mathrm{m}, 0,0)\) and has velocity \(\vec{v}_{2}=\left(-5.30 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . At this instant, what are the magnitude and direction of the magnetic force that \(q_{1}\) exerts on \(q_{2} ?\)

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 \(\mathrm{cm}\) and length \(40.0 \mathrm{cm},\) and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?
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