please help A uniform plane wave travels through free space, and its instantaneo

Question

please help

A uniform plane wave travels through free space, and its instantaneous electric field intensity vector is expressed as = 15 cos(omega t + 10 pi z + theta 0)y V/m Where time is in seconds, position z is m meters. The magnetic field intensity of the wave is H= 0.02 A/m at t = 0 and z = 1.15m. Determine The operating frequency and initial (for t = 0) phase theta 0 of the electric field in the plane Z = 0. Expression for the instantaneous magnetic field intensity vector The complex magnetic field intensity vector

Explanation / Answer

i)

| H | = | E | / 377 ohms

H = 15 cos (wt + 10 pi z + theta ) / 377

= (15/377) * cos (wt + 10 pi z + theta )

at t = 0 and z=1.15 H = 0.02

0.02 = 15/377 * cos(10*pi*1.15 + theta)

cos (10 pi *1.15 + theta ) = 0.5

11.5 pi + theta = pi / 3

theta = -35.08 degrees

for z=0

in free space

del.(E )=0

15 *10 pi*sin (wt + 10pi z + theta ) = 0

wt + 35.08 = pi (sin = 0 for 0, pi,2pi)

wt = 180-35.08 = 145

the operating frequency = w/2 pi = 145/360 = 0.4 Hz

ii) expression for instantaneous magnetic field vector

H = 15 cos (wt + 10 pi z - 35.08 ) / 377

= (15/377) * cos (wt + 10 pi z - 35.08) A/m

iii)complex magnetic field vector

direction of wave is in z-direction,direction of magnetic field isin y direction

so magnetic field intensity direction is in x - direction

H = 15 cos (wt + 10 pi z - 35.08 ) / 377

= (15/377) * cos (wt + 10 pi z - 35.08 ) x A/m

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