The graph of the magnetic energy density and the electric energy density as a function of time.

Question

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Answer 1

Question

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

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Answer 2

Question

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

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Answer 3

Question

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

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Answer 4

Question

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

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Answer 5

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

Answer 6

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)

Answer 7

Chapter 34, Problem 22PQ

To determine

The graph of the magnetic energy density and the electric energy density as a function of time.

Answer 8

Expert Solution & Answer

Answer to Problem 22PQ

The graph of electric energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 1

Figure-(1)

The graph of magnetic energy density as a function of time is as follows.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 2

Figure-(2)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The electric and magnetic fields vary in space with time $t$ . The electric field $E(x,t)$ and the magnetic field $B(x,t)$ are represented by the following equations.

$E(x,t)=Emaxsin(kx−ωt)$ (I)

$B(x,t)=Bmaxsin(kx−ωt)$ (II)

Here, $Emax$ is the amplitude of the electric field, $Bmax$ is the amplitude of the magnetic field, $k$ is the wave number and $ω$ is the angular frequency of the wave.

Write the given expression for the electric field.

$E(x,t)=3.75sin(kx−0.094t) V/m$

Compare the above expression with the equation (I).

$Emax=3.75 V/mω=0.094 rad/s$

Write the expression for the time period $T$ .

$T=2πω$

Substitute $0.094 rad/s$ for $ω$ in the above equation.

$T=2π0.094 rad/s=66.8 s$

Write the expression for electric energy density $μE$ .

$μE=12ε0[E(x,t)]2$

Here, $ε0$ is the permittivity of free space and $E(x,t)$ is the electric field.

Substitute equation (I) in the above equation and $8.85×10−12 C2N−1m−2$ for $ε0$ to find $μE$ .

$μE=12(8.85×10−12 C2N−1m−2)[3.75sin(kx−0.094t) V/m]2=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Calculate the values of $μE$ at $x=0$ for time $t=0,5 s,10 s,15 s....$

$t$ (in $s$ )	$μE$ (in $J/m3$ )
0	0
5	$1.27×10−11$
10	$4.05×10−11$
15	$6.06×10−11$
20	$5.64×10−11$
25	$3.15×10−11$
30	$6.21×10−12$
35	$1.36×10−12$
40	$2.09×10−11$
45	$4.88×10−11$
50	$6.22×10−11$
55	$5.00×10−11$
60	$2.23×10−11$
65	$1.84×10−12$
70	$5.32×10−12$

Plot the graph by taking $uE$ on the y-axis and $t$ on x-axis.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 3

Figure-(1)

Write the expression for the amplitude of the magnetic field.

$Bmax=Emaxc$

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

Substitute $3.75 V/m$ for $Emax$ and $3×108 m/s$ for $c$ in the above equation to find $Bmax$ .

$Bmax=3.75 V/m3×108 m/s=1.25×10−8 T$

Substitute $1.25×10−8 T$ for $Bmax$ and $0.094 rad/s$ for $ω$ in equation (II) to find $B(x,t)$ .

$B(x,t)=(1.25×10−8)sin(kx−0.094t) T$ (III)

Write the expression for the magnetic energy density of the wave.

$μB=[B(x,t)]22μ0$

Here, $μ0$ is the permeability of free space and $B(x,t)$ is the magnetic field.

Substitute equation (III) and $4π×10−7 T.m/A$ for $μ0$ in the above equation to find $μB$ .

$μB=[(1.25×10−8)sin(kx−0.094t)T]22(4π×10−7T.m/A)=(6.22×10−11 J/m3)sin2(kx−0.094t)$

Since, the expressions for the electric energy density and magnetic energy density are same. Therefore, the graph for magnetic energy density is same as that for electric energy density.

Hence, the graph for magnetic energy density with time is as shown.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 34, Problem 22PQ , additional homework tip 4

Figure-(2)