Topic: Electromagnetics
The current density at the surface of a thick metal plate is $100\text{ A/m}^2$. What is the skin depth if the current density at a depth of 0.0059 cm is $0.272\text{ A/m}^2$? (Assume $\ln(10) \approx 2.3$, $\ln(2.72) \approx 1$)
- A. 5 $\mu$m
- B. 10 $\mu$m
- C. 15 $\mu$m
- D. 20 $\mu$m
Answer: B. 10 $\mu$m
Explanation: Current density decays exponentially: $J = J_s e^{-\alpha z}$. Given $J_s = 100$, $J = 0.272$, and $z = 0.0059 \times 10^{-2}$ m. Then $0.272 = 100 e^{-\alpha z}$, leading to $e^{-\alpha z} = 0.00272$. Taking natural log, $-\alpha z = \ln(0.00272) = \ln(2.72) - \ln(1000) = 1 - 3(2.3) = -5.9$. Solving for $\alpha$, we get $\alpha = \frac{5.9}{0.0059 \times 10^{-2}} = 10^5$. Skin depth $\delta = 1/\alpha = 10^{-5}$ m = 10 $\mu$m.