Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 24, Problem 73P

A linearly polarized microwave of wavelength 1.50 cm is directed along the positive x axis. The electric field vector has a maximum value of 175 V/m and vibrates in the xy plane. Assuming the magnetic field component of the wave can be written in the form B = Bmax sin (kxωt), give values for (a) Bmax, (b) k, and (c) ω. (d) Determine in which plane the magnetic field vector vibrates. (e) Calculate the average value of the Poynting vector for this wave. (f) If this wave were directed at normal incidence onto a perfectly reflecting sheet, what radiation pressure would it exert? (g) What acceleration would be imparted to a 500-g sheet (perfectly reflecting and at normal incidence) with dimensions of 1.00 m × 0.750 m?

(a)

Expert Solution
Check Mark
To determine

The value for Bmax.

Answer to Problem 73P

The value for Bmax is 5.83×107T.

Explanation of Solution

Write the expression to calculate the peak value of magnetic field.

  Bmax=Emaxc

Here, Emax is the peak value for electric field, Bmax is the peak value of magnetic field and c is the speed of light.

Conclusion:

Substitute 175V/m for Emax and 3.00×108m/s for c in the above equation to calculate Bmax.

  Bmax=175V/m3.00×108m/s=5.83×107T

Thus, the value for Bmax is 5.83×107T.

(b)

Expert Solution
Check Mark
To determine

The magnitude of k.

Answer to Problem 73P

The magnitude of k is 419m1.

Explanation of Solution

Write the expression to calculate the wavenumber of k.

  k=2πλ

Here, λ is the wavelength.

Conclusion:

Substitute 1.50cm for λ to calculate k.

  k=2π1.50cm(102m1cm)=419m1

Thus, the magnitude of k is 419m1.

(c)

Expert Solution
Check Mark
To determine

The magnitude of ω.

Answer to Problem 73P

The magnitude of ω is 1.26×1011s1.

Explanation of Solution

Write the expression to calculate the angular frequency or ω.

  ω=kc

Conclusion:

Substitute 419m1 for k and 3.00×108m/s for c in the above equation to calculate ω.

  ω=(419m1)(3.00×108m/s)=1.26×1011s1

Thus, the magnitude of ω is 1.26×1011s1.

(c)

Expert Solution
Check Mark
To determine

The magnitude of ω.

Answer to Problem 73P

The magnitude of ω is 1.26×1011s1.

Explanation of Solution

Write the expression to calculate the angular frequency or ω.

  ω=kc

Conclusion:

Substitute 419m1 for k and 3.00×108m/s for c in the above equation to calculate ω.

  ω=(419m1)(3.00×108m/s)=1.26×1011s1

Thus, the magnitude of ω is 1.26×1011s1.

(d)

Expert Solution
Check Mark
To determine

The plane at which magnetic field vector vibrates.

Answer to Problem 73P

The plane of vibration of magnetic field vector is z direction.

Explanation of Solution

Here, both the electric and magnetic field vectors are vibrates in xy-plane. The direction of pointing vector is same as that of the direction of the wave since the wave carries the energy along the direction of propagation.

From the expression of the magnetic field given, the electric field vibrates along y direction. Therefore, the magnetic field must vibrate in z direction since the wave travels along the x direction which is mutually normal to both the electric and magnetic vibrations.

Conclusion:

Substitute 419m1 for k and 3.00×108m/s for c in the above equation to calculate ω.

  ω=(419m1)(3.00×108m/s)=1.26×1011s1

Thus, the plane of vibration of magnetic field vector is z direction.

(e)

Expert Solution
Check Mark
To determine

The average value of pointing vector in the wave.

Answer to Problem 73P

The average value of pointing vector is 40.6W/m2.

Explanation of Solution

Write the expression to calculate the average value of pointing vector.

  Savg=EmaxBmaxμ0

Here, Savg is the average pointing vector and μ0 is the permeability of free space.

Conclusion:

Substitute 175V/m for Emax, 4π×107Tm/A for μ0 and 5.83×107T for Bmax in the above equation to calculate Savg.

  Savg=(175V/m)(5.83×107T)4π×107Tm/A=40.6W/m2

Thus, the average value of pointing vector is 40.6W/m2.

(f)

Expert Solution
Check Mark
To determine

The radiation pressure exerted by the wave.

Answer to Problem 73P

The radiation pressure exerted by the wave is 2.71×107N/m2.

Explanation of Solution

Write the expression to calculate the radiation pressure.

  Pr=2Savgc

Here, Pr is the radiation pressure.

Conclusion:

Substitute 40.6W/m2 for Savg and 3.00×108m/s for c in the above equation to calculate Pr.

  Pr=2(40.6W/m2)3.00×108m/s=2.71×107N/m2

Thus, the radiation pressure exerted by the wave is 2.71×107N/m2.

(g)

Expert Solution
Check Mark
To determine

The acceleration imparted to the given sheet.

Answer to Problem 73P

The acceleration is 4.07×107m/s2.

Explanation of Solution

The area of the sheet is 0.750m2.

Write the expression to calculate the acceleration.

  a=PrmA

Here, a is the acceleration, A is the area and m is the mass.

Conclusion:

Substitute 2.71×107N/m2 for Pr, 0.750m2  for A and 500g for m in the above equation to calculate a.

  a=(2.71×107N/m2)500g(103kg1g)(0.750m2)=4.07×107m/s2

Thus, the acceleration is 4.07×107m/s2.

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Chapter 24 Solutions

Principles of Physics: A Calculus-Based Text

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