Let's solve the problem step by step: --- ### **1. Find the Differential Equation for \( A(x) \)** Given: \[ \vec{A} = A(x) \exp[i(\omega t - \alpha z)]\, \hat{y} \] and \[ \nabla^2 \vec{A} = \mu \tilde{\sigma} \frac{\partial \vec{A}}{\partial t} \] #### **Step 1.1: Substitute \( \vec{A} \) into the equation** Since \(\vec{A}\) only has a \(y\) component and depends on \(x, t, z\), let's write: \[ \vec{A} = \hat{y} A(x) e^{i(\omega t - \alpha z)} \] The Laplacian reduces to: \[ \nabla^2 \vec{A} = \hat{y} \left[\frac{\partial^2 A}{\partial x^2} - \alpha^2 A(x)\right] e^{i(\omega t - \alpha z)} \] The time derivative: \[ \frac{\partial \vec{A}}{\partial t} = \hat{y} (i\omega) A(x) e^{i(\omega t - \alpha z)} \] So the equation becomes: \[ \hat{y} \left[\frac{d^2 A}{dx^2} - \alpha^2 A(x)\right] e^{i(\omega t - \alpha z)} = \mu \tilde{\sigma} \hat{y} (i\omega) A(x) e^{i(\omega t - \alpha z)} \] #### **Step 1.2: Cancel common factors and simplify** Divide both sides by \(\hat{y} e^{i(\omega t - \alpha z)}\): \[ \frac{d^2 A}{dx^2} - \alpha^2 A(x) = \mu \tilde{\sigma} (i\omega) A(x) \] \[ \frac{d^2 A}{dx^2} = \left[ \alpha^2 + i\mu \tilde{\sigma} \omega \right] A(x) \] Let \( k^2 = \alpha^2 + i\mu \tilde{\sigma} \omega \): \[ \frac{d^2 A}{dx^2} - k^2 A(x) = 0 \] --- #### **Step 1.3: General solution for \(A(x)\) and apply boundary conditions** General solution: \[ A(x) = C_1 e^{k x} + C_2 e^{-k x} \] Boundary conditions: - \(A(0) = A_0\) - \(\lim_{x \to \infty} A(x) = 0\) For the solution to be finite as \(x \to \infty\), set \(C_1 = 0\): \[ A(x) = C_2 e^{-k x} \] \[ A(0) = A_0 = C_2 \implies C_2 = A_0 \] \[ \boxed{A(x) = A_0 e^{-k x}} \] where \(k = \sqrt{\alpha^2 + i\mu \tilde{\sigma} \omega}\). --- ### **2. Find \( \vec{E} \) and \( \vec{H} \), and the ratio of the \(z\) and \(x\) components of \( \vec{H} \)** #### **Step 2.1: Find \( \vec{E} \) and \( \vec{B} \) from \( \vec{A} \)** - **Electric field:** \( \vec{E} = -\frac{\partial \vec{A}}{\partial t} \) - **Magnetic field:** \( \vec{B} = \nabla \times \vec{A} \) - **Magnetic field intensity:** \( \vec{H} = \vec{B}/\mu \) Let’s compute: #### **Step 2.2: Compute \( \vec{E} \)** \[ \vec{E} = -\frac{\partial \vec{A}}{\partial t} = -i\omega A(x) e^{i(\omega t - \alpha z)} \hat{y} \] --- #### **Step 2.3: Compute \( \vec{B} = \nabla \times \vec{A} \)** \[ \vec{A} = A(x) e^{i(\omega t - \alpha z)} \hat{y} \] Calculate the curl: \[ (\nabla \times \vec{A})_x = \frac{\partial A_y}{\partial z} = -i\alpha A(x) e^{i(\omega t - \alpha z)} \] \[ (\nabla \times \vec{A})_z = -\frac{\partial A_y}{\partial x} = -\frac{dA(x)}{dx} e^{i(\omega t - \alpha z)} \] So, \[ \vec{B} = -i\alpha A(x) e^{i(\omega t - \alpha z)} \hat{x} - \frac{dA(x)}{dx} e^{i(\omega t - \alpha z)} \hat{z} \] #### **Step 2.4: Compute \( \vec{H} \)** \[ \vec{H} = \frac{\vec{B}}{\mu} \] So, \[ H_x = -\frac{i\alpha}{\mu} A(x) e^{i(\omega t - \alpha z)} \] \[ H_z = -\frac{1}{\mu} \frac{dA(x)}{dx} e^{i(\omega t - \alpha z)} \] --- #### **Step 2.5: Ratio of \(H_z\) to \(H_x\)** \[ \frac{H_z}{H_x} = \frac{-\frac{1}{\mu} \frac{dA(x)}{dx}}{-\frac{i\alpha}{\mu} A(x)} = \frac{1}{i\alpha} \frac{dA(x)}{dx} \cdot \frac{1}{A(x)} \] Given \(A(x) = A_0 e^{-kx}\): \[ \frac{dA(x)}{dx} = -k A_0 e^{-kx} = -k A(x) \] \[ \therefore \frac{H_z}{H_x} = \frac{1}{i\alpha} \frac{-k A(x)}{A(x)} = -\frac{k}{i\alpha} = \frac{ik}{\alpha} \] --- #### **Step 2.6: Express \(k\) and find modulus** Recall \(k = \sqrt{\alpha^2 + i\mu \tilde{\sigma} \omega}\). Let’s write \(k\) in terms of \(\alpha\): Let \(K = \frac{k}{\alpha} = \left(1 + \frac{i\mu \tilde{\sigma} \omega}{\alpha^2} \right)^{1/2}\) So, the modulus of \(H_z/H_x\) is: \[ \left|\frac{H_z}{H_x}\right| = \left| \frac{ik}{\alpha} \right| = |K| \] Now, for modulus: \[ |K| = \left| \left(1 + \frac{i\mu \tilde{\sigma} \omega}{\alpha^2} \right)^{1/2} \right| = \left[ \left(1 + \frac{\mu^2 \tilde{\sigma}^2 \omega^2}{\alpha^4} \right)^{1/2} \right]^{1/2} \] \[ = \left( 1 + \frac{\mu^2 \tilde{\sigma}^2 \omega^2}{\alpha^4} \right)^{1/4} \] --- ### **Final Answers** #### **1. The differential equation and solution for \(A(x)\):** \[ \frac{d^2 A}{dx^2} - (\alpha^2 + i\mu \tilde{\sigma} \omega) A(x) = 0 \] with solution (given the boundary conditions): \[ \boxed{A(x) = A_0 \exp\left[ -\sqrt{\alpha^2 + i\mu \tilde{\sigma} \omega}\ x \right]} \] #### **2. The electric and magnetic fields:** \[ \vec{E} = -i\omega A(x) e^{i(\omega t - \alpha z)} \hat{y} \] \[ \vec{H} = -\frac{i\alpha}{\mu} A(x) e^{i(\omega t - \alpha z)} \hat{x} - \frac{1}{\mu} \frac{dA(x)}{dx} e^{i(\omega t - \alpha z)} \hat{z} \] #### **3. The ratio of \(H_z\) to \(H_x\):** \[ \left| \frac{H_z}{H_x} \right| = \left( 1 + \frac{\mu^2 \tilde{\sigma}^2 \omega^2}{\alpha^4} \right)^{1/4} \] --- **This matches the required result.**