A microwave source produces pulses of 20.0-GHz radiation , with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the micro-waves into a parallel beam of radiation as shown in Figure P24.72. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

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Answer 1

Textbook Question

Chapter 24, Problem 72P

A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the micro-waves into a parallel beam of radiation as shown in Figure P24.72. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

Chapter 24, Problem 72P, A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

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Answer 2

Textbook Question

Chapter 24, Problem 72P

A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the micro-waves into a parallel beam of radiation as shown in Figure P24.72. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

Chapter 24, Problem 72P, A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

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Answer 3

Textbook Question

Chapter 24, Problem 72P

A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the micro-waves into a parallel beam of radiation as shown in Figure P24.72. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

Chapter 24, Problem 72P, A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

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Answer 4

Textbook Question

Chapter 24, Problem 72P

A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the micro-waves into a parallel beam of radiation as shown in Figure P24.72. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

Chapter 24, Problem 72P, A microwave source produces pulses of 20.0-GHz radiation, with each pulse lasting 1.00 ns. A

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

Answer 8

(a)

Expert Solution

To determine

The wavelength of the microwave.

Answer to Problem 72P

The wavelength is $0.015 m$ .

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The wavelength of the microwave.

Answer 13

Answer to Problem 72P

The wavelength is $0.015 m$ .

Answer 14

Explanation of Solution

Write the expression to calculate the wavelength.

$λ=cf$

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

Conclusion:

Substitute $20.0 GHz$ for f and $3.00×108 m/s$ for c in the above equation to calculate $λ$ .

$λ=3.00×108 m/s20.0 GHz(109 Hz1 GHz)=0.015 m$

Thus, the wavelength is $0.015 m$ .

Answer 15

(b)

Expert Solution

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The total energy of each pulse.

Answer 20

Answer to Problem 72P

The total energy is $25.0×10−6 J$ .

Answer 21

Explanation of Solution

Write the expression to calculate the total energy.

$U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0 kW$ for P and $1.00 ns$ for t in the above equation to calculate U.

$U=(25.0 kW)(103 W1 kW)(1.00 ns)(10−9 s1 ns)=25.0×10−6 J$

Thus, the total energy is $25.0×10−6 J$ .

Answer 22

(c)

Expert Solution

To determine

The average energy density inside the pulse.

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The average energy density inside the pulse.

Answer 27

Answer to Problem 72P

The average energy density is $7.37×10−3 J/m3$ .

Answer 28

Explanation of Solution

Write the expression to calculate the average energy density.

$u=Uπr2l$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

$l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

$u=Uπr2(ct)$

Conclusion:

Substitute $25.0×10−6 J$ for U, $1.00 ns$ for t, $6.00 cm$ for r and $3.00×108 m/s$ for c in the above equation to calculate u.

$u=25.0×10−6 Jπ(6.00 cm(10−2 m1 cm))2(3.00×108 m/s)(1.00 ns)(10−9 s1 ns)=7.37×10−3 J/m3$

Thus, the average energy density is $7.37×10−3 J/m3$ .

Answer 29

(d)

Expert Solution

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The amplitude of electric and magnetic field.

Answer 34

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Answer 35

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

$Emax=2uε0$ (I)

Here, $ε0$ is the permittivity of free space and $Emax$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

$Bmax=Emaxc$ (II)

Here, $Bmax$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $8.85×10−12 C2/N⋅m2$ for $ε0$ in the equation (I) to calculate $Emax$ .

$Emax=2(7.37×10−3 J/m3)8.85×10−12 C2/N⋅m2=4.08×104 V/m$

Substitute $4.08×104 V/m$ for $Emax$ and $3.00×108 m/s$ for c in the above equation (II) to calculate $Bmax$ .

$Bmax=4.08×104 V/m3.00×108 m/s=1.36×10−4 T$

Thus, the amplitude of electric and magnetic field respectively $4.08×104 V/m$ and $1.36×10−4 T$ .

Answer 36

(e)

Expert Solution

To determine

The force exerted on the surface.

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The force exerted on the surface.

Answer 41

Answer to Problem 72P

The force exerted on the surface is $8.33×10−5 N$ .

Answer 42

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

$F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

$A=πr2$

Use the above equation to rewrite the equation for F.

$F=u(πr2)$

Conclusion:

Substitute $7.37×10−3 J/m3$ for u and $6.00 cm$ for r in the above equation to calculate F.

$F=7.37×10−3 J/m3(π(6.00 cm(10−2 m1 cm))2)=8.33×10−5 N$

Thus, the force exerted on the surface is $8.33×10−5 N$ .

Answer 43

Ch. 24.1 - Prob. 24.1QQ Ch. 24.4 - Prob. 24.2QQ Ch. 24.4 - Prob. 24.3QQ Ch. 24.4 - Prob. 24.4QQ Ch. 24.6 - Prob. 24.5QQ Ch. 24.6 - Prob. 24.6QQ Ch. 24.7 - Prob. 24.7QQ Ch. 24 - Prob. 1OQ Ch. 24 - Prob. 2OQ Ch. 24 - Prob. 3OQ

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Answer to Problem 72P

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Answer to Problem 72P

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Answer to Problem 72P

Explanation of Solution

Answer to Problem 72P

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Answer to Problem 72P

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