The equations for the electric field and the magnetic field for an electromagnetic wave.

Question

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Answer 1

Question

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

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Answer 2

Question

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

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Answer 3

Question

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

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Answer 4

Question

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

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Answer 5

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Answer 6

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Answer 7

Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Answer 8

Expert Solution & Answer

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the electric field of an electromagnetic wave.

$E=Emaxsin(kx−ωt)k^$ (I)

Here, $Emax$ is the amplitude of the electric field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the magnetic field of an electromagnetic wave.

$B=Bmaxsin(kx−ωt)(−j^)$ (II)

Here, $Bmax$ is the amplitude of the magnetic field, $k$ is the propagation constant and $ω$ is the angular frequency of the wave.

Write the expression for the propagation constant.

$k=2πλ$ (III)

Here, $k$ is the propagation constant and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πcλ$ (IV)

Here, $ω$ is the angular frequency, $c$ is the speed of light and $f$ is the frequency of the wave.

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

$Bmax=Emaxc$ (V)

Conclusion:

Substitute $555 nm$ for $λ$ in (III) to find $k$ .

$k=2π(555 nm×10−9 m1 nm)=1.132×107 m−1$

Substitute, $3×108 m/s$ for $c$ and $555 nm$ for $λ$ in (IV)to find $ω$ .

$ω=2π3×108 m/s(555 nm×10−9 m−11 nm)=3.4×1015 rad/s$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $0.050 V/m$ for $Emax$ in equation(I) to find $E$ .

$E=(0.050 V/m)sin((1.132×107 m−1)x−(3.4×1015 rad/s)t)k^$ .

Substitute $3×108 m/s$ for $c$ and $0.050 V/m$ for $Emax$ in (V) to find $Bmax$ .

$Bmax=0.050 V/m3×108 m/s=1.67×10−10 T$

Substitute $1.132×107 m−1$ for $k$ , $3.4×1015 rad/s$ for $ω$ and $1.67×10−10 T$ for $Bmax$ in equation(II) to find $B$ .

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Thus, the equation for the electric field of the electromagnetic wave is given by:

$E=(0.050 V/m)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t]k^$ .

The equation for the magnetic field of the electromagnetic wave is given by:

$B=(1.67×10−10 T)sin[(1.132×107 m−1)x−(3.4×1015 rad/s)t](−j^)$ .

Answer 12

Ch. 34.1 - Prob. 34.1CE Ch. 34.2 - Prob. 34.2CE Ch. 34.4 - The electric part of an electromagnetic wave is...Ch. 34.5 - Prob. 34.4CE Ch. 34.5 - Prob. 34.5CE Ch. 34.6 - Prob. 34.6CE Ch. 34.8 - Prob. 34.7CE Ch. 34 - Prob. 1PQ Ch. 34 - Prob. 2PQ Ch. 34 - A circular coil of radius 0.50 m is placed in a...

The equations for the electric field and the magnetic field for an electromagnetic wave.

The equations for the electric field and the magnetic field for an electromagnetic wave.

Answer to Problem 32PQ

Explanation of Solution

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Chapter 34 Solutions