An electromagnetic wave frequency 10 GHz is propagating through a conductorhaving conductivity 6 x 107(Ωm)-1. Then skin depth of the conductor is ______Correct answer is '0.6'. Can you explain this answer?

Question

An electromagnetic wave frequency 10 GHz is propagating through a conductorhaving conductivity 6 x 107(&Omega;m)-1. Then skin depth of the conductor is ______Correct answer is '0.6'. Can you explain this answer?

Answer

<div style='text-align: center;'>skin depth <img src='https://cn.edurev.in/ApplicationImages/Temp/1f3a8ba8-5b40-44bc-8045-f789fb1f6ceb_lg.jpg' style='max-height: 400px;' /><br /><img src='https://cn.edurev.in/ApplicationImages/Temp/eaacb3ad-ee20-4358-ac58-282e803b7837_lg.jpg' style='max-height: 400px;' /></div>

Answer

The conductivity of a material determines how well it can conduct electricity. In this case, the conductivity of the conductor is given as 6 x 107.

To determine whether the conductor will be able to conduct the electromagnetic wave with a frequency of 10 GHz, we need to compare the frequency of the wave with the conductivity of the conductor.

The skin depth (δ) is a measure of how deeply an electromagnetic wave can penetrate into a conductor. It is given by the formula:

δ = √(2/(πμσf))

Where:
δ = skin depth
μ = permeability of the material
σ = conductivity of the material
f = frequency of the electromagnetic wave

In this case, the frequency is 10 GHz (10 x 10^9 Hz) and the conductivity is 6 x 10^7. Assuming the conductor is made of a material with a permeability of μ = 4π x 10^-7 H/m, we can calculate the skin depth:

δ = √(2/(πμσf))
= √(2/(π x 4π x 10^-7 x 6 x 10^7 x 10 x 10^9))
= √(2/(6 x 10^-7 x 6 x 10^7 x 10 x 10^9))
= √(2/(3.6 x 10^-6 x 10^16))
= √(2/(3.6 x 10^10))
≈ 1.21 x 10^-6 m

The skin depth is approximately 1.21 micrometers. This means that the electromagnetic wave with a frequency of 10 GHz will be able to penetrate into the conductor up to a depth of 1.21 micrometers.

Since the skin depth is very small compared to the size of the conductor, it is likely that the conductor will be able to conduct the electromagnetic wave effectively.

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