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
Answer: Option
Explanation:
.
Discussion:
4 comments Page 1 of 1.
Arham said:
2 years ago
@MythiliRani.
Thank you.
Thank you.
Mythili Rani said:
6 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.
= √( 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:
8 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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