Physics for Scientists and Engineers (AP Edition)
Physics for Scientists and Engineers (AP Edition)
9th Edition
ISBN: 9781133953951
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 34, Problem 34.70AP

You may wish to review Sections 16.4 and 16.8 on the transport of energy by string waves and sound. Figure P33.46 is a graphical representation of an electromagnetic wave moving in the x direction. We wish to find an expression for the intensity of this wave by means of a different process from that by which Equation 33.27 was generated. (a) Sketch a graph of the electric field in this wave at the instant t = 0, letting your flat paper represent the xy plane. (b) Compute the energy density uE in the electric field as a function of x at the instant t = 0. (c) Compute the energy density in the magnetic field uB as a function of x at that instant. (d) Find the total energy density u as a function of x, expressed in terms of only the electric field amplitude. (e) The energy in a “shoebox” of length λ and frontal area A is E λ = 0 λ u A d x . (The symbol Eλ for energy in a wavelength imitates the notation of Section 16.4.) Perform the integration to compute the amount of this energy in terms of A, λ, Emax, and universal constants. (f) We may think of the energy transport by the whole wave as a series of these shoeboxes going past as if carried on a conveyor belt. Each shoebox passes by a point in a time interval defined as the period T = 1/f of the wave. Find the power the wave carries through area A. (g) The intensity of the wave is the power per unit area through which the wave passes. Compute this intensity in terms of Emax and universal constants. (h) Explain how your result compares with that given in Equation 33.27.

Figure P33.46

Chapter 34, Problem 34.70AP, You may wish to review Sections 16.4 and 16.8 on the transport of energy by string waves and sound.

(a)

Expert Solution
Check Mark
To determine

To draw: The sketch a graph of the electric field at the instant t=0 , the flat paper representing the xy plane.

Answer to Problem 34.70AP

The graph of the electric field is shown in the diagram below.

Physics for Scientists and Engineers (AP Edition), Chapter 34, Problem 34.70AP , additional homework tip  1

Figure (1)

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ

Introduction: The electric filed is a region around a charged particle within which a force would be experienced by other particles that may be attractive force or repulsive force.

Explanation:

The electric field in the figure is perpendicular to the magnetic field and the sinusoidal wave is travelling with the speed of light. The waves move on the positive x axis and the electric field is in the xy -plane. The wavelength of the of the wave is λ .

The expression for the electric field is,

E=Emaxcos(kxωt)

Here,

Emax is the maximum value of the magnitude of electric field.

k is the angular wave number.

ω is the angular frequency.

Substitute 0 for t in the above equation to find the value of E .

E=Emaxcos(kx) (1)

The diagram of the electric field is shown below.

Physics for Scientists and Engineers (AP Edition), Chapter 34, Problem 34.70AP , additional homework tip  2

Figure (1)

(b)

Expert Solution
Check Mark
To determine
The energy density uE in the electric field as a function of x at an instant t=0 .

Answer to Problem 34.70AP

The energy density uE in the electric field as a function of x at an instant t=0 is 12εE2maxcos2(kx) .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The formula to calculate the energy density in the electric field is,

uE=12εE2

Here,

ε is the permittivity of the free space.

E is the magnitude of electric field.

Substitute Emaxcos(kx) for E in the above equation to find the value of uE .

uE=12ε(Emaxcos(kx))2=12εE2maxcos2(kx)

Thus, the energy density uE in the electric field as a function of x at an instant t=0 is 12εE2maxcos2(kx) .

Conclusion:

Therefore, the energy density uE in the electric field as a function of x at an instant t=0 is 12εE2maxcos2(kx) .

(c)

Expert Solution
Check Mark
To determine
The energy density uB in the magnetic field as a function of x at an instant t=0 .

Answer to Problem 34.70AP

The energy density uB in the magnetic field as a function of x at an instant t=0 is 12μB2maxcos2(kx) .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The expression for the magnetic field is,

B=Bmaxcos(kxωt)

Here,

Bmax is the maximum value of the magnitude of magnetic field.

k is the angular wave number.

ω is the angular frequency

Substitute 0 for t in the above equation to find the value of B .

B=Bmaxcos(kx) (2)

The formula to calculate the energy density in the magnetic is,

uB=12μB2

Here,

μ is the permissibility of the region.

B is the magnetic field.

Substitute Bmaxcos(kx) for B in the above equation to find the value of uB .

uB=12μ(Bmaxcos(kx))2=12μB2maxcos2(kx)

Thus, the energy density uB in the magnetic field as a function of x at an instant t=0 is 12μB2maxcos2(kx) .

Conclusion:

Therefore, the energy density uB in the magnetic field as a function of x at an instant t=0 is 12μB2maxcos2(kx) .

(d)

Expert Solution
Check Mark
To determine
The total energy density in terms of electric field amplitude.

Answer to Problem 34.70AP

The total energy density in terms of electric field amplitude is E2maxcos2(kx) .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The expression for the energy due to magnetic field at instant t=0 is,

uB=12μB2maxcos2(kx) (3)

The value of Bmax is given by,

Bmax=Emaxc=Emax1εμ

Substitute Emax1εμ for Bmax in equation (3) to find the value of uB .

uB=12μ(Emax1εμ)2maxcos2(kx)=12εE2maxcos2(kx)

The formula to calculate the total energy density is,

u=uE+uB

Substitute 12εE2maxcos2(kx) for uB and 12εE2maxcos2(kx) for uE in the above equation to find the value of u .

u=12εE2maxcos2(kx)+12εE2maxcos2(kx)=εE2maxcos2(kx) (4)

Thus, the total energy density in terms of electric field amplitude is εE2maxcos2(kx) .

Conclusion:

Therefore, the total energy density in terms of electric field amplitude is εE2maxcos2(kx) .

(e)

Expert Solution
Check Mark
To determine
The amount of energy in the shoebox.

Answer to Problem 34.70AP

The amount of energy in the shoebox is 12εE2maxAλ .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The expression for the energy in the shoebox is,

Eλ=0λuAdx

Here,

A is the area of the shoebox.

Substitute εE2maxcos2(kx) for u in the above equation to find the value of Eλ .

Eλ=0λεE2maxcos2(kx)Adx

Integrate the above equation to find the value of Eλ .

Eλ=εE2maxA0λ[12+12cos(2kx)]dx=12εE2maxA[x]0λ+12(2k)εE2maxA[sin(2kx)]0λ=12εE2maxAλ+14kεE2maxA[sin(4π)sin(0)]=12εE2maxAλ

Thus, the amount of energy in the shoebox is 12εE2maxAλ .

Conclusion:

Therefore, the amount of energy in the shoebox is 12εE2maxAλ .

(f)

Expert Solution
Check Mark
To determine
The power the wave carries through area A .

Answer to Problem 34.70AP

The power the wave carries through area A  is 12εcE2maxA .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The expression for the energy in the shoebox is,

Eλ=12εE2maxAλ

The formula to calculate the power is,

P=EλT (5)

Substitute 12εE2maxAλ for Eλ and 1f for T in the above equation to find value of P .

P=12εE2maxAλ1f=12εE2maxAλf=12εcE2maxA

Thus, the power the wave carries through area A  is 12εcE2maxA .

Conclusion:

Therefore, the power the wave carries through area A  is 12εcE2maxA .

(g)

Expert Solution
Check Mark
To determine
The intensity of the wave .

Answer to Problem 34.70AP

The intensity of the wave is 12εcE2max .

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The formula to calculate the intensity of the wave is,

I=PA

Substitute 12εcE2maxA for P in the above equation to find the value of I .

I=12εcE2maxAA=12εcE2max

Thus, the intensity of the wave is 12εcE2max .

Conclusion:

Therefore, the intensity of the wave is 12εcE2max .

(h)

Expert Solution
Check Mark
To determine
The comparison of the result with the equation Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

Answer to Problem 34.70AP

The expression of the result is same as that of the expression.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

Savg=EmaxBmax2μ=E2max2μc=cB2max2μ .

The formula to calculate the average intensity is,

I=12εcE2max

Substitute the average value as 1μc2 for ε in the above equation to find the average intensity.

I=121μc2cE2max=E2max2μc

Thus, the average value of the intensity is same as that given in the equation.

Conclusion:

Therefore, the expression of the result is same as that of the expression.

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Chapter 34 Solutions

Physics for Scientists and Engineers (AP Edition)

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