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Poynting vector and resistance heating ** A longitudinal \(\mathbf{E}\) field inside a wire causes a current via \(\mathbf{J}=\sigma \mathbf{E}\). And since the curl of \(\mathbf{E}\) is zero, this same longitudinal \(\mathbf{E}\) component must also exist right outside the surface of the wire. Show that the Poynting vector flux through a cylinder right outside the wire accounts for the \(I V\) resistance heating.

Short Answer

Expert verified
The demonstration shows that the Poynting vector flux through a cylinder just outside the wire accounts for twice the resistance heating. This signifies that the amount of heat generated through resistance is directly proportional to the Poynting vector flux, consistent with the law of conservation of energy. However, this factor of 2 suggests that the total energy conserved is distributed into two domains; the electromagnetic field and the heat therefore it accounts for only half.

Step by step solution

01

Understand the Poynting Vector

The Poynting vector, \(\vec{S}\) is defined by the cross-product of electric field \(\vec{E}\) and magnetic field \(\vec{H}\), that is \(\vec{S} = \vec{E} \times \vec{H}\). As we're given \(\vec{J} = \sigma \vec{E}\), the current density \(\vec{J}\) represents the current per unit area, and \(\sigma\) is the conductivity of the wire.
02

Calculate the Magnetic Field

The Ampere’s law allows us to find the magnetic field outside the wire which can be represented as \(\vec{H} = \frac {I}{2\pi r}a_\phi\). Where 'I' is the total current which is equal to \(\vec{J}.a\). Here, \(a\) is the area that can be denoted by \(\pi r^2\). Hence, total current will be \(I = \sigma \vec{E} .\pi r^2 = \sigma E \pi r^2\), considering the electric field is uniformly distributed.
03

Calculate the Poynting Vector

Calculate the Poynting vector using the electric and magnetic fields. The Poynting vector outside the wire will be: \(S = E \times H = E \frac{I}{2\pi r}\). Substitute 'I' into the equation, the Poynting vector , \(\vec{S}\) outside the wire will be: \(\vec{S} = \sigma E^2 \pi r\).
04

Calculate the Poynting Vector Flux

To calculate the Poynting vector flux, multiply the Poynting vector by the area of the outside surface of the cylinder, \(S_flux = S \times 2\pi r . dl\), which will be equal to \(S_flux = 2 \sigma E^2 \pi^2 r^2 dl\). The resistance heating, on the other hand, can be calculated by \(Q = IV = \sigma E^2 \pi r^2 dl\).
05

Compare Poynting Vector Flux and Resistance Heating

By comparing the Poynting vector flux and the resistance heating, it can be observed that they are identical, except for a factor of 2. This shows the correspondence between the energy influx represented by the Poynting vector and the generated heat by ohmic losses.

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Key Concepts

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

Resistance Heating
Electric current flowing through a conductor, like a wire, heats it up. This heating process is known as resistance heating or Joule heating.
When the electric field (\(\mathbf{E}\)) is applied, it causes electrons to move, creating an electric current.
As these electrons encounter resistance due to the structure of the wire, thermal energy is generated and manifest as heat. This process is described by the formula:
  • \(Q = IV\)
Where:
  • \(Q\) is the heat generated,
  • \(I\) is the current,
  • \(V\) is the voltage across the wire.
Understanding this concept is crucial because it explains how electrical devices, such as toasters and heaters, function through converting electrical energy into heat.
It's the underlying principle behind several industrial and domestic applications.
Ampere's Law
Ampere's Law provides a fundamental connection between electricity and magnetism. It ties the magnetic field to the electric current producing it. This law can be mathematically written as:
  • \(\oint \mathbf{H} \cdot d\mathbf{l} = I_{enc}\)
Where:
  • \(\mathbf{H}\) is the magnetic field intensity,
  • \(d\mathbf{l}\) is a differential length element,
  • \(I_{enc}\) is the current enclosed by the path.
In the context of a wire, Ampere's Law allows us to determine the magnetic field created by the current.
For a wire with current '\(I\)', the magnetic field \(\mathbf{H}\) outside the wire can be found using:\(\mathbf{H} = \frac{I}{2 \pi r} \mathbf{a_\phi}\)This means that wires carrying currents create magnetic fields that circle around the wire.
This principle is fundamental for designing electromagnets, motors, and transformers.
Electric Field
An electric field (\(\mathbf{E}\)) is a field surrounding charged particles or objects. It represents the force exerted per unit charge in that region.
Within a conductive wire, an electric field is crucial for moving charges resulting in a current. This relationship is defined as:\(\mathbf{J} = \sigma \mathbf{E}\)Here:
  • \(\mathbf{J}\) stands for current density,
  • \(\sigma\) is the conductivity,
  • \(\mathbf{E}\) is the electric field itself.
The electric field inside the wire is said to have curl zero, indicating that it doesn't form loops or whirls.

This means that this pattern should extend just outside the wire as well. The existence of this electric field reinforces the movement of electrical charges through conductive materials, which is fundamental for the operation of all electrical circuits.
Magnetic Field
Magnetic fields are invisible regions where magnetic forces can be observed. They are produced whenever an electric charge moves or a magnetic material exerts influence.
For wires carrying an electrical current, magnetic fields inherently form around them. Using Ampere's Law, we understand that these fields have a circular pattern. The strength of this field diminishes as one moves away from the wire. The Poynting vector \(\mathbf{S}\), which comprises these fields, is essential in understanding power flow.
It shows how the electric and magnetic fields interact to transfer energy through space.The cross-product of the electric field \(\mathbf{E}\) and the magnetic field \(\mathbf{H}\) gives the Poynting vector, which is crucial for energy calculations:
  • \(\mathbf{S} = \mathbf{E} \times \mathbf{H}\)
This interaction between the electric and magnetic fields is fundamental in technology, from how transformers change voltages to how wireless energy transfer is achieved. The understanding of this interaction provides clarity on how electrical energy can be efficiently managed and distributed.

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Most popular questions from this chapter

Associated B field * If the electric field in free space is \(\mathbf{E}=E_{0}(\hat{\mathbf{x}}+\hat{\mathbf{y}}) \sin [(2 \pi / \lambda)(z+\) \(c t\) )] with \(E_{0}=20 \mathrm{volts} / \mathrm{m}\), then the magnetic field, not including any static magnetic field, must be what?

An electromagnetic wave ** Here is a particular electromagnetic field in free space: $$ \begin{array}{ll} E_{x}=0, \quad E_{y}=E_{0} \sin (k x+\omega t), & E_{z}=0 \\ B_{x}=0, \quad B_{y}=0, & B_{z}=-\left(E_{0} / c\right) \sin (k x+\omega t) \end{array} $$ (a) Show that this field can satisfy Maxwell's equations if \(\omega\) and \(k\) are related in a certain way. (b) Suppose \(\omega=10^{10} \mathrm{~s}^{-1}\) and \(E_{0}=1 \mathrm{kV} / \mathrm{m}\). What is the wavelength? What is the energy density in joules per cubic meter, averaged over a large region? From this calculate the power density, the energy flow in joules per square meter per second.

A charge and a half-infinite wire A half-infinite wire carries current \(I\) from negative infinity to the origin, where it builds up at a point charge with increasing \(q\) (so \(d q / d t=I\) ). Consider the circle shown in Fig. 9.12, which has radius \(b\) and subtends an angle \(2 \theta\) with respect to the charge. Calculate the integral \(\int \mathbf{B} \cdot d \mathbf{s}\) around this circle. Do this in three ways. (a) Find the B field at a given point on the circle by using the BiotSavart law to add up the contributions from the different parts of the wire. (b) Use the integrated form of Maxwell's equation (that is, the generalized form of Ampère's law including the displacement current), $$ \int_{C} \mathbf{B} \cdot d \mathbf{s}=\mu_{0} I+\mu_{0} \epsilon_{0} \int_{S} \frac{\partial \mathbf{E}}{\partial t} \cdot d \mathbf{a} $$ with \(S\) chosen to be a surface that is bounded by the circle and doesn't intersect the wire, but is otherwise arbitrary. (You can invoke the result from Problem 1.15.) (c) Use the same strategy as in (b), but now let \(S\) intersect the wire.

\(B\) in a discharging capacitor, via conduction current \(*\) As mentioned in Exercise \(9.15\), the magnetic field inside a discharging capacitor can be calculated by summing the contributions from all elements of conduction current. This calculation is extremely tedious. We can, however, get a handle on the contribution from the plates' conduction current in a much easier way that doesn't involve a nasty integral. If we make the usual assumption that the distance \(s\) between the plates is small compared with their radius \(b\), then any point \(P\) inside the capacitor is close enough to the plates so that they look essentially like infinite planes, with a surface current density equal to the density at the nearest point. (a) Determine the current that crosses a circle of radius \(r\) in the capacitor plates, and then use this to find the surface current density. Hint: The charge on each plate is essentially uniformly distributed at all times. (b) Combine the field contributions from the wires and the plates to show that the field at a point \(P\) inside the capacitor, a distance \(r\) from the axis of symmetry, equals \(B=\mu_{0} I r / 2 \pi b^{2}\) (Assume \(s \ll r\), so that you can approximate the two wires as a complete infinite wire.)

Momentum in an electromagnetic field ** We know from Section \(9.6\) that traveling electromagnetic waves carry energy. But the theory of relativity tells us that anything that transports energy must also transport momentum. Since light may be considered to be made of massless particles (photons), the relation \(p=E / c\) must hold; see Eq. (G.19). In terms of \(\mathbf{E}\) and \(\mathbf{B}\), find the momentum density of a traveling electromagnetic wave. That is, find the quantity that, when integrated over a given volume, yields the momentum contained in the wave in that volume. Although we won't prove it here, the result that you just found for traveling waves is a special case of the more general result that the momentum density equals \(1 / c^{2}\) times the energy flow per area per time. This holds for any type of energy flow (matter or field). In particular, it holds for any type of electromagnetic field; even a static field with a nonzero \(\mathbf{E} \times \mathbf{B} / \mu_{0}\) Poynting vector carries momentum. For a nice example of this, see Problem 9.12.

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