Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
Question
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Chapter 34, Problem 77PQ
To determine

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

Expert Solution & Answer
Check Mark

Answer to Problem 77PQ

The equation for the electric field is E=(1.44×104V/m)sin[(1.01×107m1)z(3.02×1015rad/s)t] and the equation for the magnetic field is B=(48.0×106T)sin[(1.01×107m1)z(3.02×1015rad/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(kxωt)V/m                                                          (V)

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

    BZ=Bmaxsin(kxωt)T                                                           (VI)

Conclusion:

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

    f=3×108m/s(625nm×109m1nm)

  =4.8×1014Hz

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

    ω=2π(4.8×1014)=3.01×1015rad/s

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

    k=2π(625nm×109m1nm)=1.01×107m1

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

    Emax=(3.0×108m/s)(48.0μT×109T1μT)=1.44×104V/m

Substitute 1.44×104V/m for Emax, 1.01×107m1 for k, and 3.01×1015rad/s for ω in the (V) equation to find EZ

    EZ=(1.44×104V/m)sin[(1.01×107m1)z(3.02×1015rad/s)t]

Substitute 48.0μT for Bmax, 1.01×107m1 for k, and 3.01×1015rad/s for ω in(VI) equation to find BZ

    EZ=(1.01×106T)sin[(1.01×107m1)z(3.02×1015rad/s)t]

Therefore, the equation for the electric field is E=(1.44×104V/m)sin[(1.01×107m1)z(3.02×1015rad/s)t] and the equation for the magnetic field is B=(48.0×106T)sin[(1.01×107m1)z(3.02×1015rad/s)t].

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Chapter 34 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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