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At 50 MHz a lossy dielectric material is characterized by μ = 2.1μo ,ε = 3.6 εo and σ = 0.08 S/m. The electric field is Es = 6e-jγx uz V/m
Que: The propagation constant γ is
  • a)
    7.43 + j2.46 per meter
  • b)
    2.46 + j7.43 per meter
  • c)
    6.13 + j5.41 per meter
  • d)
    5.41 + j6.13 per meter
Correct answer is option 'D'. Can you explain this answer?
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Calculation of Propagation Constant:

Given parameters:
- Frequency, f = 50 MHz
- Dielectric constant, εr = 2.1
- Loss tangent, tan(δ) = 3.6o
- Conductivity, σ = 0.08 S/m
- Electric field, Es = 6e-jxuz V/m

First, we need to calculate the loss angle, δ:
- tan(δ) = 3.6o
- δ = tan^-1(3.6o) = 0.063 radians

Next, we can calculate the complex permittivity, ε:
- ε = εr - j tan(δ) = 2.1 - j 0.012

Then, we can calculate the complex propagation constant, γ:
- γ = α + j β
- α = σ/2 ε ε0 ω = 0.00022 Np/m (where ε0 is the permittivity of free space and ω is the angular frequency)
- β = ω √(μ ε) = 35.81 Np/m (where μ is the permeability of the medium)

Finally, we can calculate the total propagation constant, γtot:
- γtot = α + j β = 5.41 + j 6.13 Np/m

Answer: The correct answer is option 'D' which represents the total propagation constant as 5.41 + j 6.13 Np/m.
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At 50 MHz a lossy dielectric material is characterized byμ = 2.1μo,ε = 3.6εoandσ = 0.08 S/m.The electric field is Es= 6e-jγxuzV/mQue: The propagation constantγ isa)7.43 + j2.46 per meterb)2.46 + j7.43 per meterc)6.13 + j5.41 per meterd)5.41 + j6.13 per meterCorrect answer is option 'D'. Can you explain this answer?
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