Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 34, Problem 16P
To determine

The verification for E=Emaxcos(kxωt) and B=Bmaxcos(kxωt) being the solution to equations 34.15 and 34.16 respectively.

Expert Solution & Answer
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Answer to Problem 16P

The verification for E=Emaxcos(kxωt) and B=Bmaxcos(kxωt) being the solution to equations 34.15 and 34.16 respectively is as stated below.

Explanation of Solution

The given equations are.

    E=Emaxcos(kxωt)                                                                                                 (I)

Here, E is the magnitude of electric field , Emax is the magnitude of the maximum electric field, k is the wave vector and ω is the angular frequency.

    B=Bmaxcos(kxωt)                                                                                               (II)

Here, B is the magnitude of magnetic field , Bmax is the magnitude of the maximum magnetic field.

Differentiate equation (I) partially twice with respect to x.

    Ex=Emaxsin(kxωt)(k)2Ex2=Emaxcos(kxωt)(k2)                                                                                (III)

Differentiate equation (I) partially twice with respect to t.

    Et=Emaxsin(kxωt)(ω)2Et2=Emaxcos(kxωt)(ω)2                                                                               (IV)

Differentiate equation (II) partially twice with respect to x.

    Bx=Bmaxsin(kxωt)(k)2Bx2=Bmaxcos(kxωt)(k2)                                                                                 (V)

Differentiate equation (II) partially twice with respect to t.

    Bt=Bmaxsin(kxωt)(ω)2Bt2=Bmaxcos(kxωt)(ω)2                                                                               (VI)

Divide equation (III) by equation (IV).

    2Ex22Et2=Emaxcos(kxωt)(k2)Emaxcos(kxωt)(ω)22Ex2=k2ω2(2Et2)                                                                          (VII)

Divide equation (V) by equation (VI).

    2Bx22Bt2=Bmaxcos(kxωt)(k2)Bmaxcos(kxωt)(ω)22Ex2=k2ω2(2Bt2)                                                                         (VIII)

Also,

    kω=μ0ε0                                                                                                              (IX)

Here, μ0 is the permeability of the medium and ε0 is the permittivity if the medium.

Substitute μ0ε0 for kω in equation (VII).

    2Ex2=(μ0ε0)2(2Et2)2Ex2=μ0ε0(2Et2)                                                                                           (X)

The equation 34.15 is ,

    2Ex2=μ0ε0(2Et2)                                                                                                 (XI)

Substitute μ0ε0 for kω in equation (VIII).

    2Bx2=(μ0ε0)2(2Bt2)2Bx2=μ0ε0(2Bt2)                                                                                          (XII)

The equation 34.16 is,

    2Bx2=μ0ε0(2Bt2)                                                                                              (XIII)

Conclusion:

Equation (X) is equal to equation (XI) and equation (XII) equals to equal (XIII).

Hence, verified.

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Chapter 34 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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