hw-1-soln - − 0.8 π × 0.75 φ mA m ⇒ φ = 153.6 ◦...

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7.7 For f=60 MHz, ϵ r =4, μ r =1, k = ω ϵ r c = 2 π × 6 × 10 7 × 2 3 × 10 8 = 0.8 π [rad/m] E is polarized along ˆ z direction, wave travels along ˆ x E ( x , t ) = ˆ zE 0 cos ( 2 π × 6 × 10 7 t + 0.8 π x + φ ) [V/m] Note that the + sign indicates the wave travels in ˆ x η = η 0 ϵ r = 120 π 2 = 60 π [ ] , H = 1 η ( ˆ x ) × E , ( ˆ x × ˆ z = ˆ y ) Thus, H ( x , t ) = ˆ y E 0 η cos ( 2 π × 10 7 t + 0.8 π x + φ ) [A/m] Since E 0 η = 10[mA/m], E 0 = 10 × 60 π × 10 3 = 0.6 π [V/m]. H ( 0.75 m , 0 ) = 7 [ mA / m ] = 10 cos
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Unformatted text preview: ( − 0.8 π × 0.75 + φ )[ mA / m ] ⇒ φ = 153.6 ◦ Thus, ( ⃗ E ( x , t ) = ˆ z · 0.6 π cos ( 1.2 π × 10 8 t + 0.8 π x + 153.6 ◦ ) [V/m] ⃗ H ( x , t ) = ˆ y · 10 − 2 cos ( 1.2 π × 10 8 t + 0.8 π x + 153.6 ◦ ) [A/m] 7.9 7.11 7.13...
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