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Chapter 8: Problem 4

A long straight wire of radius a carries a current \(I\) and has resistance \(R_{1}\) per unit length. Compute \(\mathbf{E}\) and \(\mathbf{B}\) at its surface and show that the rate of energy flow into the wire via the Poynting flux is \(I^{2} R_{1}\) per unit length.

Short Answer

Expert verified
Answer: The electric field at the surface of the wire is given by \(E = \frac{I}{\pi a^2 \sigma_{1}}\), and the magnetic field is given by \(B = \frac{\mu_{0}I}{2\pi a}\). The rate of energy flow into the wire is calculated through the Poynting vector, and the Poynting flux demonstrates that the rate of energy flow into the wire is equal to \(I^{2}R_{1}\) per unit length, where \(R_{1}\) represents the resistance per unit length.

Step by step solution

01

Find the magnetic field at the surface of the wire using Ampere's Law

We will use Ampere's Law, which is given by: \(\oint_{C} \mathbf{B} \cdot d\mathbf{l} = \mu_{0}I_{\text{enclosed}}\) Since the current is uniformly distributed over the wire, we can write the current density as \(j = \frac{I}{\pi a^2}\). We choose an Amperian loop as a circle of radius \(r\) which is the same as the radius of the wire (\(r = a\)). The magnetic field \(\mathbf{B}\) is tangential to the circle and constant in magnitude. Therefore, the dot product simplifies to: \(B \oint_{C} dl = B \cdot 2\pi a\) The enclosed current within the loop is given by multiplying the current density by the area enclosed by the loop: \(I_{\text{enclosed}} = j \cdot \pi a^2 = I\) Applying Ampere's Law and solving for the magnetic field \(B\): \(B \cdot 2\pi a = \mu_{0}I\) \(B = \frac{\mu_{0}I}{2\pi a}\) This is the magnetic field at the surface of the wire.
02

Find the electric field inside the wire using Ohm's Law

Inside the wire, there will be an electric field due to the potential difference and the resistance of the wire. Ohm's law relates the current, the electric field, and the resistance as follows: \(I = E \cdot A \cdot \sigma_{1}\) where \(\sigma_{1}\) is the conductivity of the wire and \(A = \pi a^2\) is the cross-sectional area of the wire. From Ohm's law, the electric field inside the wire with radius \(a\) is given by: \(E = \frac{I}{\pi a^2 \sigma_{1}}\)
03

Compute the Poynting vector

The Poynting vector \(\mathbf{S}\) represents the power flow per unit area in an electromagnetic field and is given by the cross product of electric and magnetic fields: \(\mathbf{S} = \frac{1}{\mu_{0}}\mathbf{E} \times \mathbf{B}\) The electric field is radially inward, and the magnetic field is counterclockwise tangent to the wire's surface. Thus the Poynting vector is directed into the wire. The magnitude of \(\mathbf{S}\) at the surface of the wire can be written as: \(|\mathbf{S}| = \frac{1}{\mu_{0}} E B = \frac{1}{\mu_{0}} \left( \frac{I}{\pi a^2 \sigma_{1}}\right)\left(\frac{\mu_{0}I}{2\pi a}\right) = \frac{I^2}{2\pi a^3 \sigma_{1}}\)
04

Calculate the rate of energy flow into the wire per unit length

To find the rate of energy flow into the wire per unit length, we integrate the Poynting vector over the wire's surface: \(\int |\mathbf{S}| dA = \int_{0}^{2\pi} \int_{0}^{a} \frac{I^2}{2\pi a^3 \sigma_{1}} \cdot a d\theta dr =\) \(\left[\frac{I^2}{2 a^2 \sigma_{1}} \int_{0}^{a} dr \right] \int_{0}^{2\pi} d\theta = \frac{I^2}{2 a^2 \sigma_{1}} \cdot a \cdot 2\pi = I^{2} R_{1}\) where \(R_{1} = \frac{1}{\sigma_{1}}\) is the resistance per unit length. This gives the rate of energy flow into the wire via the Poynting flux, proving that it is equal to \(I^{2}R_{1}\) per unit length.

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Key Concepts

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

Understanding Ampere's Law
Ampere's Law is fundamental to understanding the relationship between electric currents and magnetic fields. At its core, Ampere's Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability of the vacuum times the electric current enclosed by the loop.

This concept helps us to deduce the magnetic field surrounding an electric current. Applied to a straight wire, Ampere's Law helps us calculate the field at various positions relative to the wire. This is particularly useful in the case of a wire carrying a uniform current. When we construct an imaginary loop (Amperian loop) parallel to the wire, we can measure the circulation of the magnetic field around this loop to determine the field's strength at the surface of the wire.

In the problem we're considering, once the current and radius of the wire are known, the law simplifies the process to a straightforward calculation by avoiding the complexities of the internal structure of the wire. With Ampere's Law, we could efficiently solve step 1 of the problem and find the magnetic field at the wire's surface, which is vital for later steps involving the Poynting vector.
Ohm's Law's Application in Electric Fields
Ohm's Law is a critical principle in electrical engineering and physics, connecting an electric current that flows through a conductor with the voltage across it and the resistance the conductor provides. The foundation of Ohm's Law is the equation \( V = IR \), which states that the voltage (\( V \)) across a conductor is equal to the product of the current (\( I \)) flowing through it and the resistance (\( R \)) of it.

In the context of our problem, Ohm's Law proves useful for finding the electric field within the wire. Since resistance is directly related to the material's resistivity and inversely to the area, for any section of the wire, the internal electric field helps drive the current across that resistance. A key point to remember is that resistance here is not the conventional resistance but resistance per unit length, as represented by \( R_1 \).

Using Ohm's Law, we determined the electric field in step 2 inside the wire, which is later used to calculate the Poynting vector. It shows the importance of electrical properties like conductivity in determining the characteristics of the electromagnetic fields in conductive materials.
The Significance of the Poynting Vector in Energy Flow
The Poynting vector, named after John Henry Poynting, is a pivotal concept in electromagnetism that represents the energy transfer per unit area in an electromagnetic field. It is defined as the cross product of the electric field and magnetic field vectors.

Mathematically, \( \mathbf{S} = \frac{1}{\mu_{0}} \mathbf{E} \times \mathbf{B} \), where \( \mathbf{S} \) is the Poynting vector, \( \mathbf{E} \) is the electric field, \( \mathbf{B} \) is the magnetic field, and \( \mu_{0} \) is the permeability of free space. In our problem, the Poynting vector tells us how much electromagnetic power flows into the wire per unit area. By integrating this over the surface of the wire, we can find the total power flow or the rate of energy transfer into the wire per unit length.

The significance of the Poynting vector in this context is two-fold: it not only quantifies the rate of energy transfer but also offers directionality to the flow. As such, in step 3, we saw how the cross product of \( \mathbf{E} \) and \( \mathbf{B} \) directed the energy flow into the wire, showcasing how electromagnetic fields interact to transfer energy. Ultimately, the exercise demonstrates the energy flow into the wire is consistent with the power dissipation given by \( I^2R_1 \), where \( R_1 \) is the resistance per unit length, thus connecting the Poynting vector to real-world electrical measurements.

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Most popular questions from this chapter

Consider total reflection at an interface between two nonmagnetic media, with relative refractive index \(n=c_{1} / c_{2}<1\). For angles of incidence \(\theta_{1}\) exceeding the critical angle of (8.6.42), Snell's law gives $$ \sin \theta_{2}=\frac{\sin \theta_{1}}{n}>1, $$ which implies that \(\theta_{2}\) is a complex angle with an imaginary cosine, $$ \cos \theta_{2}=\left(1-\sin ^{2} \theta_{2}\right)^{1 / 2}=j\left(\frac{\sin ^{2} \theta_{1}}{n^{2}}-1\right)^{1 / 2} $$ Substitute these relations in the case I reflection coefficient (8.6.28) to establish $$ R_{\mathbf{E} \perp}=e^{-i 2 \phi_{\perp}}, $$ where $$ \tan \phi_{\perp}=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / x}}{\cos \theta_{1}} $$ That is, the magnitude of the reflection coefficient is unity, but the phase of the reflected wave depends upon angle. Similarly show for case II from (8.6.36), that \(R_{\text {III }}=e^{-\text {jod with }}\) $$ \tan \phi \|=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / 2}}{n^{2} \cos \theta_{1}}=\frac{1}{n^{2}} \tan \phi_{\perp} . $$ Note that the two phase shifts are different, so that in general the state of polarization of an incident wave is altered.

Consider an electric dipole consisting of a charge \(-e\) oscillating sinusoidally in position about a stationary charge \(+e\). Show that the instantaneous total power radiated can be written in the form $$ \frac{d W}{d t}=\frac{e^{2}[a]^{2}}{6 \pi \epsilon_{0} c^{3}} $$ where \([a]=a_{0} e^{i(\omega t-k n)}\) is the instantaneous (retarded) acceleration of the moving charge Since this result does not depend upon the oscillator frequency, and since by Fourier analysis, an arbitrary motion can be described by superposing many sinusoidal motions of proper frequency, amplitude, and phase, this rate-of-radiation formula has general validity for any accelerated charge (in the nonrelativistic limit \(\diamond \ll c\) ).

Develop Poynting's theorem for the general material medium of relative permittivity \(\kappa_{\&}\) and permeability \(\kappa_{m}\) introduced in Prob. \(8.2 .2\); i.e., substitute the Maxwell curl equations \((8.2 .18)\) and \((8.2 .20)\) in the expansion of \(\nabla \cdot(\mathbf{E} \times \mathbf{H})\) to obtain $$ \oint_{S}(\mathbf{E} \times \mathbf{H}) \cdot d \mathbf{S}+\int_{V}\left(\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}+\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}\right) d v+\int_{V} \mathbf{E} \cdot \mathbf{J} d v=0 $$ from which it follows that the Poynting vector is $$ s=\mathbf{E} \times \mathbf{H} $$ and the energy density is $$ \begin{aligned} \left(W_{1}\right)_{\text {leld }} &=\frac{1}{2} \mathbf{E} \cdot \mathbf{D}+\frac{1}{2} \mathbf{H} \cdot \mathbf{B} \\ &=\frac{1}{2} \kappa_{\ell} \epsilon_{0} E^{2}+\frac{1}{2 \kappa_{\mathrm{m}} \mu_{0}} B^{2} \end{aligned} $$ What restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\) are necessary to obtain \((8.4 .24)\) ?

Consider an inhomogeneous dielectric medium, i.e., one for which the dielectric constant is a function of position, \(\kappa_{e}=\kappa_{e}(x, y, z)\). Show that the fields obey the wave equations $$ \begin{aligned} &\nabla^{2} \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ &\nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}}=-\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$ where, in general, the terms on the right-hand sides couple the cartesian components of the fields. Now introduce the special case that the permittivity changes only in the direction of propagation (the \(z\) direction, say) and show that for monochromatic plane waves the equations become $$ \begin{aligned} &\frac{d^{2} \mathbf{E}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{\theta}(z) \mathbf{E}=0 \\ &\frac{d^{2} \mathbf{B}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{e}(z) \mathbf{B}=\frac{1}{\kappa_{e}(z)} \frac{d x_{e}}{d z} \frac{d \mathbf{B}}{d z} \end{aligned} $$ Approximate solution of this type of equation is discussed in Sec. \(9.1 .\)

Consider two unbounded plane waves whose vector wave numbers \(k_{1}\) and \(\kappa_{2}\left(\left.\right|_{1,2} \mid=\right.\) \(\omega / c)\) define a plane and whose electric fields are polarized normal to the plane. (a) Show that the superposition of these two plane waves is a wave traveling in the direction bisecting the angle \(\alpha\) between \({ }_{1}{ }_{1}\) and \({ }_{k}\) and that the \(\mathbf{E}\) field vanishes on a set of nodal planes spaced \(a=\) \(\lambda_{0} / 2 \sin \frac{1}{2} \alpha\) apart. \((b)\) Show that plane conducting walls can be placed at two adjacent nodal planes without violating the electromagnetic boundary conditions and likewise that a second pair of conducting walls of arbitrary separation \(b\) can be introduced to construct a rectangular waveguide of cross section \(a\) by \(b\), propagating the \(T E_{10}\) mode. Thus establish that the TE \(_{10}\) mode (more generally, the TE \(_{10}\) modes) may be interpreted as the superposition of two plane waves making the angle \(\frac{1}{2} \alpha\) with the waveguide axis and undergoing multiple reflections from the sidewalls. Note: The situation is directly analogous to that discussed in Sec. 2.4. Higherorder TE modes \((m>0)\) and TM modes may be described similarly as a superposition of four plane waves.
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