Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 3 (Q.No. 47)
47.
A 300 MHz plane EM wave is propagating in free space. The wave is incident normally on an infinite copper slap. The antenuation constant for the wave is (σcopper = 5.8 x 107 mho/ m). The skin depth is
0.005 x 10-3 m
0.0038 x 10-3m
0.0028 x 10-3 m
0.0024 x 10-3 m
Answer: Option
Explanation:

.

Discussion:
4 comments Page 1 of 1.

Arham said:   2 years ago
@MythiliRani.

Thank you.

Mythili Rani said:   5 years ago
The skin depth = √( 2 / (ω*μ*σ )),
= √( 2 / ( 2 * 3.14 * 300 * 10^6 * 4 * 3.14 * 10^7 * 5.8 * 10 ^ 7)),
= 0.0038 * 10^-3.
where
μ = permeability (4π* 10-7 H/m).
ω = radian frequency = 2π*f (Hz).

GIVEN
σ = 5.8 x 107 mho/ m.
f=300 MHz.

Vinay said:   7 years ago
Why 4 * 3.14 * 10^-7?

Amnesh said:   8 years ago
Skin depth = 1/sqr (3.14 * 300 * 10^6 * 4 * 3.14 * 10^_7 * 5.8 * 10^7).

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