Physics for Scientists and Engineers, Volume 1, Chapters 1-22
Physics for Scientists and Engineers, Volume 1, Chapters 1-22
8th Edition
ISBN: 9781439048382
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 34, Problem 34.60AP

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 microwaves into a parallel beam of radiation as shown in Figure P34.60. 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 34, Problem 34.60AP, A microwave source produces pulses of 20.0GHz radiation, with each pulse lasting 1.00 ns. A

Figure P34.60

(a)

Expert Solution
Check Mark
To determine

The wavelength of microwaves.

Answer to Problem 34.60AP

The wavelength of microwaves is 0.015m .

Explanation of Solution

Given Info: The frequency of pulse is 20.0GHz , time duration of each pulse is 1.00ns , radius of parabolic reflector is 6.00cm and the average power during each pulse is 25.0kW .

Formula to calculate the wavelength is,

λ=cf

Here,

f is the frequency of pulse.

c is the speed of waves.

Substitute 20.0GHz for f and 3×108m/s for c in above equation.

λ=3×108m/s20.0GHz=3×108m/s20.0×109Hz=0.015m

Conclusion:

Therefore, the wavelength of microwaves is 0.015m .

(b)

Expert Solution
Check Mark
To determine

The total energy contained in each pulse.

Answer to Problem 34.60AP

The total energy contained in each pulse is 2.5×105J .

Explanation of Solution

Given Info: The frequency of pulse is 20.0GHz , time duration of each pulse is 1.00ns , radius of parabolic reflector is 6.00cm and the average power during each pulse is 25.0kW .

Formula to calculate the total energy is,

TE=Pavg×t

Here,

Pavg is the average power.

t is the time duration of each pulse.

Substitute 25.0kW for Pavg and 1.00ns for t in the above equation.

TE=25.0kW×1.00ns=25.0×103W×1.00×109s=25×106J=2.5×105J

Conclusion:

Therefore, the total energy contained in each pulse is 2.5×105J .

(c)

Expert Solution
Check Mark
To determine

The average energy density inside each pulse.

Answer to Problem 34.60AP

The average energy density inside each pulse is 7.37×103J/m3 .

Explanation of Solution

Given Info: The frequency of pulse is 20.0GHz , time duration of each pulse is 1.00ns , radius of parabolic reflector is 6.00cm and the average power during each pulse is 25.0kW .

Formula to calculate the average energy density for each pulse is,

ug=Ic=PavgAc=Pavgπr2c

Here,

Pavg is the average power.

r is the radius.

c is the velocity of pulse.

A is the cross sectional area.

Substitute 25.0kW for Pavg , 3.14 for π , 6.00cm for r and 3×108m/s for c in the above equation.

ug=25.0kW3.14×(6.00cm)2×(3×108m/s)=25.0×103W3.14×(6.00×102m)2×(3×108m/s)=7.37×103J/m3

Conclusion:

Therefore, the average energy density inside each pulse is 7.37×103J/m3 .

(d)

Expert Solution
Check Mark
To determine

The amplitude of electric and magnetic fields in the microwaves.

Answer to Problem 34.60AP

The amplitude of electric field is 40.8×103V/m and amplitude of magnetic field is 1.36×104T .

Explanation of Solution

Given Info: The frequency of pulse is 20.0GHz , time duration of each pulse is 1.00ns , radius of parabolic reflector is 6.00cm and the average power during each pulse is 25.0kW .

Formula to calculate the magnitude of electric field is,

Emax=2ugεo

Here,

ug is the average energy density for each pulse.

Substitute 7.37×103J/m3 for ug and 8.85×1012C2/Nm2 for εo in the above equation.

Emax=2×7.37×103J/m38.85×1012C2/Nm2=40810.98V/m=40.8×103V/m

Formula to calculate the magnitude of magnetic field is,

Bmax=Emaxc

Here,

Emax is the electric field magnitude.

c is the velocity of microwave.

Substitute for 40.8×103V/m Emax and 3×108m/s for c in the above equation.

Bmax=40.8×103V/m3×108m/s=1.36×104T

Conclusion:

Therefore, the amplitude of electric field is 40.8×103V/m and amplitude of magnetic field is 1.36×104T .

(e)

Expert Solution
Check Mark
To determine

The force exerted on the surface on each pulse.

Answer to Problem 34.60AP

The force exerted on the surface on each pulse is 8.33×105N .

Explanation of Solution

Given Info: The frequency of pulse is 20.0GHz , time duration of each pulse is 1.00ns , radius of parabolic reflector is 6.00cm and the average power during each pulse is 25.0kW .

Formula to calculate the force exerted is,

F=ugA=ugπr2

Here,

ug is the average energy density of each pulse.

A is the cross sectional area.

r is the radius of pulse.

Substitute 7.37×103J/m3 for ug , 3.14 for π and 6.00cm for r in the above equation.

F=(7.37×103J/m3)×3.14×(6.00cm)2=(7.37×103J/m3)×3.14×(6.00×102m)2=8.33×105N

Conclusion:

Therefore, the force exerted on the surface on each pulse is 8.33×105N .

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Chapter 34 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

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