Question
A uniform plane wave of 100 kHz travelling in free space strikes a large block of a material having
ε = 4 ε0 , µ = 9µ0 and σ = 0 normal to the surface. If the incident
magnetic field vector is given by
B = 10-6 cos (ωt − βy) zˆ tesla
write the complete expressions for the incident, reflected, and transmitted field vectors.
Answer :
Word Count : 751
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Skip in 10
Let's solve this step by step manually. We'll deal with a uniform plane wave incident normally on a dielectric material. --- ### Given: * Frequency: $f = 100\ \text{kHz} = 10^5\ \text{Hz}$ * Free-space: $\epsilon_0, \mu_0$ * Material: $\epsilon = 4 \epsilon_0, \mu = 9 \mu_0, \sigma = 0$ * Incident magnetic field: $$ \mathbf{B}_i = 10^{-6} \cos (\omega t - \beta y) \hat{z} \quad \text{T} $$ * Wave propagating along y, normal incidence. --- ### Step 1: Find free-space wave parameters Angular frequency: $$ \omega = 2 \pi f = 2 \pi \cdot 10^5 = 6.283 \times 10^5 \ \text{rad/s} $$ Wave number in free space: $$ \beta_0 = \omega \sqrt{\mu_0 \epsilon_0} = \frac{\omega}{c} $$ Where $c = 3 \times 10^8 \ \text{m/s}$. So: $$ \beta_0 = \frac{6.283 \times 10^5}{3 \times 10^8} \approx 2.094 \times 10^{-3}\ \text{rad/m} $$ Wave impedance in free space: $$ \eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 120\pi \approx 377\ \Omega $$ --- ### Step 2: Find wave parameters in material * Material is lossless ($\sigma = 0$). * Relative permittivity: $\epsilon_r = 4$ * Relative permeability: $\mu_r = 9$ Wave number in material: $$ \beta = \omega \sqrt{\mu \epsilon} = \omega \sqrt{9 \mu_0 \cdot 4 \epsilon_0} = \omega \cdot 6
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