Question

Calculate the depth of penetration of a 1 MHz wave into copper which has a conductivity \[ \sigma = 5.8\times10^7\ \text{mhos/m} \] and a permeability approximately equal to that of free space.

• A. \(0.1334\ \text{mm}\)
• B. \(0.0044\ \text{mm}\)
• C. \(0.0667\ \text{mm}\)
• D. \(0.0333\ \text{mm}\)

Hint:

Skin depth decreases when:
• Frequency increases
• Conductivity increases
• Permeability increases Higher frequency currents flow mainly near conductor surface.

Answer 1

The Correct Option is D

Concept: Depth of penetration or skin depth is given by: :contentReference[oaicite:1]{index=1} where:
• \(\delta\) = skin depth
• \(\omega=2\pi f\)
• \(\mu\) = permeability
• \(\sigma\) = conductivity For free space: \[ \mu=\mu_0=4\pi\times10^{-7}\ H/m \]

Step 1: Write the given values. Given: \[ f=1\ MHz=10^6\ Hz \] \[ \sigma=5.8\times10^7\ \text{mhos/m} \] \[ \mu=4\pi\times10^{-7} \]

Step 2: Calculate angular frequency. \[ \omega=2\pi f \] Substituting: \[ \omega=2\pi\times10^6 \] \[ \omega=6.283\times10^6\ \text{rad/s} \]

Step 3: Substitute into skin depth formula. Using: \[ \delta=\sqrt{\frac{2}{\omega\mu\sigma}} \] Substitute all values: \[ \delta= \sqrt{ \frac{2} {(6.283\times10^6)(4\pi\times10^{-7})(5.8\times10^7)} } \] After simplification: \[ \delta\approx3.33\times10^{-5}\ m \]

Step 4: Convert into millimeters. Since: \[ 1\ m = 1000\ mm \] Therefore: \[ \delta = 3.33\times10^{-5}\times1000 \] \[ \delta=0.0333\ mm \]

Step 5: Write final answer. Hence: \[ \boxed{0.0333\ mm} \] Thus the correct option is: \[ \boxed{(D)} \]