/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A very large flat sheet carries ... [FREE SOLUTION] | 91Ó°ÊÓ

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A very large flat sheet carries a uniformly distributed electric current with current per unit width \(J_{s} .\) Example 30.6 demonstrated that the current creates a magnetic field on both sides of the sheet, parallel to the sheet and perpendicular to the current, with magnitude \(B=\frac{1}{2} \mu_{0} J_{s} .\) If the current oscillates in time according to $$ \mathbf{J}_{s}=J_{\max }(\cos \omega t) \hat{\mathbf{j}}=J_{\max }[\cos (-\omega t)] \hat{\mathbf{j}} $$ the sheet radiates an electromagnetic wave as shown in Figure P34.37. The magnetic field of the wave is described by the wave function \(\quad \mathbf{B}=\frac{1}{2} \mu_{0} J_{\max }[\cos (k x-\omega t)] \hat{\mathbf{k}}\) (a) Find the wave function for the electric field in the wave. (b) Find the Poynting vector as a function of \(x\) and \(t\) (c) Find the intensity of the wave. (d) What If? If the sheet is to emit radiation in each direction (normal to the plane of the sheet) with intensity \(570 \mathrm{W} / \mathrm{m}^{2},\) what maximum value of sinusoidal current density is required?

Short Answer

Expert verified
The electric field wave function is \( E = (1/2) \mu_0 J_max c sin(kx - \omega t) \). The Poynting vector is \( S = (1/4) \mu_0 J_max^2 c sin(2kx - 2\omega t) \). The intensity of the wave is \( I = (1/4) \mu_0 J_max^2 c \). If the radiation occurs with an intensity of 570 W/m^2, the maximum value of sinusoidal current density required is \( J_max= \sqrt{(4 \times 570) / (4\pi \times 10^{-7} \times 3.00 \times 10^8)} \, A/m \).

Step by step solution

01

Find the electric field wave function

We can apply Faraday's law to find the electric field from the given magnetic field. Faraday's law says that \( \nabla \times E = -\partial B/\partial t \). We transform this to \( E = -\partial B/(\partial x \partial t) \). Substituting the given magnetic field \( B = (1/2) \mu_0 J_max cos(kx - \omega t) \), we obtain \( E = ((1/2) k \mu_0 J_max \omega sin(kx - \omega t))/(\omega c) \). The \( k \) cancels with \( \omega/c \), resulting in \( E = (1/2) \mu_0 J_max c sin(kx - \omega t) \), where \( c \) is the speed of light.
02

Find the Poynting vector

The Poynting vector \( S \) conveys the direction and magnitude of electromagnetic energy flow and is defined as the cross product of the electric field \( E \) and the magnetic intensity \( H \), as \( S = E \times H \). We can write \( H = B/\mu_0 \) using the given magnetic field, and substitute this and the electric field we computed in step 1 into the formula for \( S \), obtaining \( S = ((1/2) \mu_0 J_max c sin(kx - \omega t))((1/2) J_max cos(kx - \omega t))/(\mu_0) \). This simplifies to \( S = (1/4) \mu_0 J_max^2 c sin(2kx - 2\omega t) \).
03

Find the intensity of the wave

The intensity of the wave is equal to the peak value of the Poynting vector. The peak of the sinusoidal function \( sin(2kx - 2\omega t) \) occurs at \( \pm 1 \). Therefore, the peak value of our Poynting vector is \( S = (1/4) \mu_0 J_max^2 c \). We refer this as the intensity \ (I\) of the wave, thus, \( I = (1/4) \mu_0 J_max^2 c \).
04

Compute the maximum value of sinusoidal current density

For the sheet to emit radiation with \( I = 570 \, W/m^2 \), we can find the maximum value of the sinusoidal current density \( J_max \) by rearranging the formula for intensity: \( J_max = \sqrt{(4I) / (\mu_0 c)} \). Substituting \( \mu_0 = 4\pi \times 10^{-7} \, T\cdot m/A \), \( c = 3.00 \times 10^8 \, m/s \), and \( I = 570 \, W/m^2 \), we find \( J_max = \sqrt{(4 \times 570) / (4\pi \times 10^{-7} \times 3.00 \times 10^8)} \, A/m \).

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Key Concepts

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

Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle in physics that explains how electric fields are produced by changing magnetic fields. In simple terms, it can be expressed with Maxwell's equation: \( abla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \). This means that the curl of the electric field \( \mathbf{E} \) is equal to the negative rate of change of the magnetic field \( \mathbf{B} \) over time.

In the context of electromagnetic waves, Faraday's Law is critical as it allows us to determine the electric field from a time-varying magnetic field. For the given problem, we start by considering the magnetic field wave function provided: \( \mathbf{B} = \frac{1}{2} \mu_0 J_{\max} \cos(k x - \omega t) \). Applying Faraday's Law allows us to derive the electric field wave function as \( E = \frac{1}{2} \mu_0 J_{\max} c \sin(kx - \omega t) \), where \( c \) is the speed of light.

Faraday's Law, thus, plays a vital role in connecting changing magnetic fields to resulting electric fields, enabling us to understand the dynamic relationship between the two in electromagnetic waves.
Poynting Vector
The Poynting vector is a crucial concept when studying electromagnetic waves, representing the power flow or energy transfer per unit area of the wave. Mathematically, it is defined as the cross product of the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{H} = \mathbf{B}/\mu_0 \):
  • \( \mathbf{S} = \mathbf{E} \times \mathbf{H} \)
This vector gives us both the direction and magnitude of electromagnetic energy flow.

By using the electric field and magnetic field expressions derived earlier, we can compute the Poynting vector as:
  • \( S = \left(\frac{1}{4}\right) \mu_0 J_{\max}^2 c \sin(2kx - 2\omega t) \)
This equation suggests that the Poynting vector oscillates spatially and temporally, capturing the wave nature of electromagnetic energy flow. Notably, the intensity of the electromagnetic wave, which is the average power per unit area, can be found by averaging the Poynting vector across a full wave cycle.

Understanding the Poynting vector helps visualize how electromagnetic waves carry energy from one place to another, which is fundamental in many applications, including telecommunications and power transmission.
Electric Field Wave Function
The electric field wave function describes how the electric field component of an electromagnetic wave varies in space and time. It's important to understand this function to analyze the properties of electromagnetic waves, like their speed, frequency, and wavelength.

In the exercise, the electric field wave function is derived using Faraday's Law from the provided magnetic field wave equation. The resulting wave function is:
  • \( E(x,t) = \frac{1}{2} \mu_0 J_{\max} c \sin(kx - \omega t) \)
Here, \( \omega \) is the angular frequency, \( k \) is the wave number, and \( c \) is the speed of light. This function tells us how the electric field vector moves and changes in time and space within the wave, oscillating as a sinusoidal function just like the related magnetic field.

The electric field is crucial in transmitting energy across space. It determines how the wave interacts with charged particles, enabling electromagnetic waves to exert forces, create heat, or transfer information via signal waves. Understanding how to derive and interpret the electric field wave function is vital in the broader study of electromagnetic theory.

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Most popular questions from this chapter

A 10.0 -mW laser has a beam diameter of \(1.60 \mathrm{mm}\). (a) What is the intensity of the light, assuming it is uniform across the circular beam? (b) What is the average energy density of the beam?

Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phase with each other. In which directions are (a) the strongest and (b) the weakest signals radiated?

In a region of free space the electric field at an instant of time is \(\mathbf{E}=(80.0 \hat{\mathbf{i}}+32.0 \hat{\mathbf{j}}-64.0 \hat{\mathbf{k}}), \mathrm{N} / \mathrm{C}\) and the magnetic field is \(\mathbf{B}=(0.200 \hat{\mathbf{i}}+0.0800 \hat{\mathbf{j}}+0.290 \hat{\mathbf{k}}) \mu \mathrm{T}\) (a) Show that the two fields are perpendicular to each other. (b) Determine the Poynting vector for these fields.

The intensity of solar radiation at the top of the Earth's atmosphere is \(1340 \mathrm{W} / \mathrm{m}^{2} .\) Assuming that \(60 \%\) of the incoming solar energy reaches the Earth's surface and assuming that you absorb \(50 \%\) of the incident energy, make an order-of-magnitude estimate of the amount of solar energy you absorb in a \(60-\) min sunbath.

Consider a small, spherical particle of radius \(r\) located in space a distance \(R\) from the Sun. (a) Show that the ratio \(F_{\mathrm{rad}} / F_{\mathrm{grav}}\) is proportional to \(1 / r,\) where \(F_{\mathrm{rad}}\) is the force exerted by solar radiation and \(F_{\mathrm{grav}}\) is the force of gravitational attraction. (b) The result of part (a) means that, for a sufficiently small value of \(r,\) the force exerted on the particle by solar radiation exceeds the force of gravitational attraction. Calculate the value of \(r\) for which the particle is in equilibrium under the two forces. (Assume that the particle has a perfectly absorbing surface and a mass density of \(1.50 \mathrm{g} / \mathrm{cm}^{3} .\) Let the particle be located \(3.75 \times 10^{11} \mathrm{m}\) from the Sun, and use \(214 \mathrm{W} / \mathrm{m}^{2}\) as the value of the solar intensity at that point.)

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