Question

29. The electric field of a sinusoidal electromagnetic wave in vacuum is given: $E_x = E_y = 0$, $E_z = E_0 \cos(ky - \omega t)$. The magnetic field of this wave is A. $B_x = B_y = 0$, $B_z = B_0 \times \cos(ky - \omega t)$ B. $B_x = B_y = 0$, $B_z = B_0 \times \sin(ky - \omega t)$ C. $B_y = B_z = 0$, $B_x = B_0 \times \cos(ky - \omega t)$ D. $B_y = B_z = 0$, $B_x = B_0 \times \cos(ky + \omega t)$ E. $B_x = B_z = 0$, $B_y = B_0 \times \cos(ky + \omega t)$

          29. The electric field of a sinusoidal electromagnetic wave in vacuum is given: $E_x = E_y = 0$,
$E_z = E_0 \cos(ky - \omega t)$. The magnetic field of this wave is
A. $B_x = B_y = 0$, $B_z = B_0 \times \cos(ky - \omega t)$
B. $B_x = B_y = 0$, $B_z = B_0 \times \sin(ky - \omega t)$
C. $B_y = B_z = 0$, $B_x = B_0 \times \cos(ky - \omega t)$
D. $B_y = B_z = 0$, $B_x = B_0 \times \cos(ky + \omega t)$
E. $B_x = B_z = 0$, $B_y = B_0 \times \cos(ky + \omega t)$

29. The electric field of a sinusoidal electromagnetic wave in vacuum is given: Ex = Ey = 0,
Ez = E0 cos(ky - ω t). The magnetic field of this wave is
A. Bx = By = 0, Bz = B0 ×cos(ky - ω t)
B. Bx = By = 0, Bz = B0 ×sin(ky - ω t)
C. By = Bz = 0, Bx = B0 ×cos(ky - ω t)
D. By = Bz = 0, Bx = B0 ×cos(ky + ω t)
E. Bx = Bz = 0, By = B0 ×cos(ky + ω t)

Added by Sara D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ajay Singhal

09/02/2021

Step 1

Step 1: The electric field of the wave is given by $E_z = E_0 cos(ky - \omega t)$. Show more…

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29. The electric field of a sinusoidal electromagnetic wave in vacuum is given: $E_x = E_y = 0$, $E_z = E_0 cos(ky - \omega t)$. The magnetic field of this wave is A. $B_x = B_y = 0$, $B_z = B_0 \times cos(ky - \omega t)$ B. $B_x = B_y = 0$, $B_z = B_0 \times sin(ky - \omega t)$ C. $B_y = B_z = 0$, $B_x = B_0 \times cos(ky - \omega t)$ D. $B_y = B_z = 0$, $B_x = B_0 \times cos(ky + \omega t)$ E. $B_x = B_z = 0$, $B_y = B_0 \times cos(ky + \omega t)$