The equations for the electric field and the magnetic field of the wave.

Question

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Answer 1

Question

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

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Answer 2

Question

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

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Answer 3

Question

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

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Answer 4

Question

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

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Answer 5

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Answer 6

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Answer 7

Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Answer 8

Expert Solution & Answer

Answer to Problem 77PQ

The equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

$f=cλ$ (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $λ$ is the wavelength.

Write the expression for the angular frequency of the wave.

$ω=2πf$ (II)

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

Write the expression for the wave number of the wave.

$k=2πλ$ (III)

Here, $k$ is the wave number and $λ$ is the wavelength of the wave.

Write the expression for the maximum electric field.

$Emax=cBmax$ (IV)

Here, $Emax$ is the maximum electric field, $c$ is the speed of light and $Bmax$ is the maximum magnetic field.

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

$EZ=Emaxsin(kx−ωt) V/m$ (V)

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

$BZ=Bmaxsin(kx−ωt) T$ (VI)

Conclusion:

Substitute $3×108 m/s$ for $c$ and $625 nm$ for $λ$ in equation (I) to find $f$ .

$f=3×108 m/s(625 nm×10−9 m1 nm)$

$=4.8×1014 Hz$

Substitute equation (I) in the equation (II) to find $ω$ .

$ω=2π(4.8×1014)=3.01×1015 rad/s$

Substitute $625 nm$ for $λ$ in the (III)equation to find $k$ .

$k=2π(625 nm×10−9 m1 nm)=1.01×107 m−1$

Substitute $48.0 μT$ for $Bmax$ and $3×108 m/s$ for $c$ in (IV) to find $Emax$ .

$Emax=(3.0×108 m/s)(48.0 μT×10−9 T1 μT)=1.44×104 V/m$

Substitute $1.44×104 V/m$ for $Emax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in the (V) equation to find $EZ$

$EZ=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Substitute $48.0 μT$ for $Bmax$ , $1.01×107 m−1$ for $k$ , and $3.01×1015 rad/s$ for $ω$ in(VI) equation to find $BZ$

$EZ=(1.01×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$

Therefore, the equation for the electric field is $E=(1.44×104 V/m)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ and the equation for the magnetic field is $B=(48.0×10−6 T)sin[(1.01×107 m−1)z−(3.02×1015 rad/s)t]$ .

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 of the wave.

The equations for the electric field and the magnetic field of the wave.

Answer to Problem 77PQ

Explanation of Solution

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Chapter 34 Solutions