Let's solve each part **step-by-step** to find the **amplitude of the displacement current density** for each scenario. --- ## **Background** The displacement current density is defined as: \[ \mathbf{J}_d = \frac{\partial \mathbf{D}}{\partial t} \] where \(\mathbf{D} = \epsilon \mathbf{E}\). \[ J_d = \epsilon \frac{\partial E}{\partial t} \] For a sinusoidal field: \[ E = E_0 \cos(\omega t - kx) \] \[ \frac{\partial E}{\partial t} = -E_0 \omega \sin(\omega t - kx) \implies \text{amplitude is } E_0 \omega \] So, \[ (J_d)_{\text{max}} = \epsilon E_0 \omega \] --- ## **(a) Adjacent to an automobile antenna** Given: \[ H_x = 0.15 \cos[3.12(3 \times 10^8 t - y)]\, \text{A/m} \] **Step 1:** Find angular frequency \(\omega\): \[ \cos[3.12(3 \times 10^8 t - y)] = \cos[\omega t - k y] \] So, \[ \omega = 3.12 \times 3 \times 10^8 = 9.36 \times 10^8\, \text{rad/s} \] **Step 2:** Relate \(H\) to \(E\) in free space: \[ E = \eta H \quad (\eta = 377\, \Omega) \] So, \[ E_x = 377 \times 0.15 \cos(\ldots) = 56.55 \cos(\ldots)\, \text{V/m} \] So, \(E_0 = 56.55\, \text{V/m}\). **Step 3:** Find \(J_d\) amplitude: \[ J_{d, \text{max}} = \epsilon_0 E_0 \omega \] \[ \epsilon_0 = 8.854 \times 10^{-12}\, \text{F/m} \] \[ J_{d,\text{max}} = 8.854 \times 10^{-12} \times 56.55 \times 9.36 \times 10^8 \] **Step 4:** Calculate: First, \(56.55 \times 9.36 \times 10^8 = 529.308 \times 10^8 = 5.29308 \times 10^{10}\) Now, \[ J_{d,\text{max}} = 8.854 \times 10^{-12} \times 5.29308 \times 10^{10} = 8.854 \times 5.29308 \times 10^{-2} \] \[ = 46.84 \times 10^{-2} = 0.468\, \text{A/m}^2 \] **Final Answer (a):** \[ \boxed{0.468\, \text{A/m}^2} \] --- ## **(b) In the air space at a point within a large power distribution transformer** Given: \[ B = 0.8 \cos[1.257 \times 10^{-6}(3 \times 10^8 t - x)]\, a_y\, \text{T} \] **Step 1:** Find \(\omega\): \[ 1.257 \times 10^{-6} \times 3 \times 10^8 = 1.257 \times 3 \times 10^2 = 3.771 \times 10^{2} = 377.1\, \text{rad/s} \] So, \(\omega = 377.1\, \text{rad/s}\) **Step 2:** Find \(E_0\): For a plane wave in air: \[ E = cB \] \[ E_0 = c B_0 = 3 \times 10^8 \times 0.8 = 2.4 \times 10^8\, \text{V/m} \] But this is enormous—physically, in transformers, the fields are not plane waves. But let's proceed as per the question's form. Alternatively, for a sinusoidal time-varying field, displacement current density can also be calculated by: \[ J_{d,\text{max}} = \epsilon_0 \omega E_0 \] But since only \(B\) is given, let's relate \(\frac{\partial B}{\partial t}\) to \(J_d\): But as per Maxwell's equations, in free space, \[ \frac{\partial D}{\partial t} = \nabla \times H \] But for a wave, the amplitude of displacement current density is: \[ J_{d,\text{max}} = \epsilon_0 E_0 \omega \] Let’s use \(E_0 = c B_0\), so \(E_0 = 3 \times 10^8 \times 0.8 = 2.4 \times 10^8\, \text{V/m}\). **Step 3:** Plug in values: \[ J_{d,\text{max}} = 8.854 \times 10^{-12} \times 2.4 \times 10^8 \times 377.1 \] First, \(2.4 \times 10^8 \times 377.1 = 9.0504 \times 10^{10}\) \[ J_{d,\text{max}} = 8.854 \times 10^{-12} \times 9.0504 \times 10^{10} = 8.854 \times 9.0504 \times 10^{-2} = 80.09 \times 10^{-2} = 0.8009\, \text{A/m}^2 \] **Final Answer (b):** \[ \boxed{0.800\, \text{A/m}^2} \] --- ## **(c) Within a large, oil-filled power capacitor** Given: - Relative permittivity \( \epsilon_r = 5 \) - \( E = 0.9 \cos[1.257 \times 10^{-6}(3 \times 10^8 t - z/\sqrt{5})]\, a_x\, \text{MV/m} \) Express \( E_0 = 0.9\, \text{MV/m} = 0.9 \times 10^6\, \text{V/m} \) **Step 1:** Find \(\omega\): \[ 1.257 \times 10^{-6} \times 3 \times 10^8 = 3.771 \times 10^2 = 377.1\, \text{rad/s} \] **Step 2:** Permittivity: \[ \epsilon = \epsilon_0 \epsilon_r = 8.854 \times 10^{-12} \times 5 = 4.427 \times 10^{-11}\, \text{F/m} \] **Step 3:** Calculate \(J_{d,\text{max}}\): \[ J_{d,\text{max}} = \epsilon E_0 \omega = 4.427 \times 10^{-11} \times 0.9 \times 10^6 \times 377.1 \] First, \(0.9 \times 10^6 \times 377.1 = 339,390\) \[ J_{d,\text{max}} = 4.427 \times 10^{-11} \times 339,390 = 1.502 \times 10^{-5} \] But, let's check the math: \(4.427 \times 10^{-11} \times 0.9 \times 10^6 = 4.427 \times 0.9 \times 10^{-11 + 6} = 3.9843 \times 10^{-5}\) Then, \(\times 377.1 = 3.9843 \times 10^{-5} \times 377.1 = 0.01502\) So, \[ J_{d,\text{max}} \approx 0.0150\, \text{A/m}^2 \] **Final Answer (c):** \[ \boxed{0.0150\, \text{A/m}^2} \] --- ## **(d) In a metallic conductor at 60 Hz** Given: - \( \epsilon = 0 \) - \( \mu = \mu_0 \) - \( \sigma = 5.8 \times 10^7\, \text{S/m} \) - \( J = \sin(377t - 117.1z)\, a_x\, \text{MA/m}^2 \) We are to find the **displacement current density amplitude**. But in a metallic conductor, **\(\epsilon = 0\)**, so **displacement current density is zero**. But the question seems to want the value: - \(J = \sin(\ldots)\) in **MA/m\(^2\)**, so amplitude is \(1 \times 10^6\) A/m\(^2\). But displacement current is: \[ J_d = \epsilon \frac{\partial E}{\partial t} \] But \(\epsilon = 0\), so \(J_d = 0\). However, perhaps the question wants the small **but nonzero** displacement current in a good conductor. Let’s suppose the displacement current density is: \[ J_{d, \text{max}} = \epsilon_0 \omega E_0 \] But from \(J = \sigma E\): \[ E_0 = \frac{J_0}{\sigma} \] So, \[ J_{d, \text{max}} = \epsilon_0 \omega \frac{J_0}{\sigma} \] Given: - \(J_0 = 1\, \text{MA/m}^2 = 1 \times 10^6\, \text{A/m}^2\) - \(\omega = 2\pi f = 2\pi \times 60 = 377\, \text{rad/s}\) - \(\sigma = 5.8 \times 10^7\, \text{S/m}\) - \(\epsilon_0 = 8.854 \times 10^{-12}\, \text{F/m}\) Plug in: \[ J_{d,\text{max}} = 8.854 \times 10^{-12} \times 377 \times \frac{1 \times 10^6}{5.8 \times 10^7} \] Calculate denominator: \[ \frac{1 \times 10^6}{5.8 \times 10^7} = \frac{1}{58} \approx 0.01724 \] Now, \[ 8.854 \times 10^{-12} \times 377 = 3.340 \times 10^{-9} \] \[ 3.340 \times 10^{-9} \times 0.01724 = 5.76 \times 10^{-11} \] Convert to pA/m\(^2\): \(1\, \text{pA} = 10^{-12}\, \text{A}\), so: \[ 5.76 \times 10^{-11}\, \text{A/m}^2 = 57.6\, \text{pA/m}^2 \] **Final Answer (d):** \[ \boxed{57.6\, \text{pA/m}^2} \] --- ## **Summary Table** | Part | Displacement Current Density Amplitude | |------|---------------------------------------| | (a) | \(0.468\, \text{A/m}^2\) | | (b) | \(0.800\, \text{A/m}^2\) | | (c) | \(0.0150\, \text{A/m}^2\) | | (d) | \(57.6\, \text{pA/m}^2\) | --- **These match the answers you provided.** --- ### **Boxed Final Answers** \[ \boxed{ \begin{aligned} (a) &\quad 0.468\, \text{A/m}^2 \\ (b) &\quad 0.800\, \text{A/m}^2 \\ (c) &\quad 0.0150\, \text{A/m}^2 \\ (d) &\quad 57.6\, \text{pA/m}^2 \\ \end{aligned} } \]