Let's break down the problem step by step: --- ### **Step 1: Given Data** - Incident magnetic field in air: \(\mathbf{H}_i(x, t) = 4\sin(\omega t - 5x)\ \mathbf{u}_y\) (A/m) - Plastic region: - Relative permeability: \(\mu_r = 1\) - Relative permittivity: \(\epsilon_r = 4\) - Conductivity: \(\sigma = 0\) (so it's a perfect dielectric) --- ## **(a) Obtain the Total Electric Field in Air** ### **Step 2: Write the Incident and Reflected Fields** Let the interface be at \(x = 0\), air is for \(x < 0\), plastic for \(x > 0\). - **Incident wave** (traveling \(+\hat{x}\)): \[ \mathbf{H}_i(x, t) = H_{0i} \sin(\omega t - k_1 x)\ \mathbf{u}_y \] \( H_{0i} = 4 \) \( k_1 = 5 \) (from \(\sin(\omega t - 5x)\)), so \[ \mathbf{H}_i(x, t) = 4\sin(\omega t - 5x)\ \mathbf{u}_y \] - **Reflected wave** (traveling \(-\hat{x}\)): \[ \mathbf{H}_r(x, t) = H_{0r} \sin(\omega t + k_1 x)\ \mathbf{u}_y \] ### **Step 3: Electric Field using Maxwell's Equations** In a plane wave in air (\(\mathbf{E} \perp \mathbf{H}\)), use: \[ \mathbf{E} = \eta \mathbf{H} \times \mathbf{\hat{k}} \] where \(\eta\) is the intrinsic impedance. - In air: \(\eta_1 = \eta_0 = 377\ \Omega\) Both waves propagate in the \(x\)-direction, so the electric field will be in the \(\mathbf{u}_z\) direction (right-hand rule). #### **Incident Electric Field:** \[ \mathbf{E}_i(x, t) = \eta_1 H_{0i} \sin(\omega t - k_1 x) \mathbf{u}_z \] \[ = 377 \times 4 \sin(\omega t - 5x) \mathbf{u}_z = 1508 \sin(\omega t - 5x) \mathbf{u}_z\ (\text{V/m}) \] #### **Reflected Electric Field:** For the reflected wave, \[ \mathbf{E}_r(x, t) = -\eta_1 H_{0r} \sin(\omega t + k_1 x) \mathbf{u}_z \] The negative sign arises because the direction of propagation is opposite. ### **Step 4: Boundary Conditions and Reflection Coefficient** At \(x = 0\), the tangential \(E\) and \(H\) fields must be continuous. - **Reflection coefficient for normal incidence:** \[ \Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1} \] where \(\eta_2\) is the impedance in plastic. - In plastic: - \(\epsilon_2 = \epsilon_r \epsilon_0 = 4\epsilon_0\) - \(\mu_2 = \mu_0\) - \(\eta_2 = \sqrt{\frac{\mu_2}{\epsilon_2}} = \frac{\eta_0}{2} = \frac{377}{2} = 188.5\ \Omega\) So, \[ \Gamma = \frac{188.5 - 377}{188.5 + 377} = \frac{-188.5}{565.5} = -0.333 \] Thus, \[ H_{0r} = \Gamma H_{0i} = -0.333 \times 4 = -1.332 \] ### **Step 5: Total Electric Field in Air** \[ \mathbf{E}_{\text{air}}(x, t) = \mathbf{E}_i(x, t) + \mathbf{E}_r(x, t) \] \[ = 1508 \sin(\omega t - 5x)\ \mathbf{u}_z - 377 \times 1.332 \sin(\omega t + 5x)\ \mathbf{u}_z \] \[ = 1508 \sin(\omega t - 5x)\ \mathbf{u}_z - 502.4 \sin(\omega t + 5x)\ \mathbf{u}_z \] **Or, explicitly:** \[ \boxed{ \mathbf{E}_{\text{total}}(x, t) = 1508 \sin(\omega t - 5x)\ \mathbf{u}_z - 502.4 \sin(\omega t + 5x)\ \mathbf{u}_z\ \text{(V/m)} } \] --- ## **(b) Calculate the Time-Average Power Density in the Plastic Region** ### **Step 6: Transmitted Wave in Plastic** - Transmission coefficient for magnetic field: \[ T = 1 + \Gamma = 1 - 0.333 = 0.667 \] \[ H_{0t} = T \times H_{0i} = 0.667 \times 4 = 2.668 \] - In plastic, wavenumber \(k_2 = k_1/\sqrt{\epsilon_r} = 5/2 = 2.5\ \text{(since } k = \omega \sqrt{\mu\epsilon}\text{)}\). - Intrinsic impedance \(\eta_2 = 188.5\ \Omega\). - **Transmitted electric field:** \[ E_{0t} = \eta_2 H_{0t} = 188.5 \times 2.668 = 503.3\ \text{V/m} \] ### **Step 7: Average Power Density** The time-average (Poynting vector) for a wave: \[ \langle S \rangle = \frac{1}{2} E_0 H_0 \] So, \[ \langle S \rangle = \frac{1}{2} \times 503.3 \times 2.668 = 671.6\ \text{W/m}^2 \] --- ## **Final Answers** ### **(a) Total Electric Field in Air** \[ \boxed{ \mathbf{E}_{\text{total}}(x, t) = 1508 \sin(\omega t - 5x)\ \mathbf{u}_z - 502.4 \sin(\omega t + 5x)\ \mathbf{u}_z\ \text{(V/m)} } \] ### **(b) Time-Average Power Density in Plastic** \[ \boxed{ \langle S \rangle = 671.6\ \text{W/m}^2 } \] --- **Let me know if you'd like the phasor (complex exponential) form or further steps for clarification!**