The electric field intensity E(x,t) =10 sin (ωt-5x) ay +10cos(ωt- 5x t+βx)az V/m, sketch wave for ωt=0,π/4, π/2, 3π/4 and π in x=0 plane and examine the direction of propagation, polarization of wave
Question
The electric field intensity E(x,t) =10 sin (ωt-5x) ay +10cos(ωt- 5x t+βx)az V/m, sketch wave for ωt=0,π/4, π/2, 3π/4 and π in x=0 plane and examine the direction of propagation, polarization of wave
Solution
To sketch the wave for the given values of ωt, we need to substitute these values into the equation for E(x,t) and plot the resulting functions.
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For ωt=0, the electric field intensity E(x,0) = 10 sin(-5x) ay + 10 cos(βx) az. This is a sinusoidal wave in the y direction and a cosine wave in the z direction.
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For ωt=π/4, E(x,π/4) = 10 sin(π/4 - 5x) ay + 10 cos(π/4 - 5x + βx) az. This is a shifted sinusoidal wave in the y direction and a shifted cosine wave in the z direction.
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For ωt=π/2, E(x,π/2) = 10 sin(π/2 - 5x) ay + 10 cos(π/2 - 5x + βx) az. This is another shifted sinusoidal wave in the y direction and a shifted cosine wave in the z direction.
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For ωt=3π/4, E(x,3π/4) = 10 sin(3π/4 - 5x) ay + 10 cos(3π/4 - 5x + βx) az. This is yet another shifted sinusoidal wave in the y direction and a shifted cosine wave in the z direction.
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For ωt=π, E(x,π) = 10 sin(π - 5x) ay + 10 cos(π - 5x + βx) az. This is a sinusoidal wave in the y direction that has been shifted by π, and a cosine wave in the z direction that has also been shifted by π.
The direction of propagation of the wave can be determined by looking at the sign of the x coefficient in the argument of the sine and cosine functions. In this case, the wave is propagating in the negative x direction.
The polarization of the wave can be determined by the direction of the electric field. In this case, the electric field has components in both the y and z directions, so the wave is polarized in the yz plane.
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