Consider a medium such that μ=μ0,ε = 81ε0,σ = 20S/m. The frequency at which the conduction current density is 10 times the displacement current density in magnitude is ____(GHz)Correct answer is '36'. Can you explain this answer?

Question

Answer

on putting all these values.

= 36 GHz

Answer

Given parameters:
- μ = μ0
- ε = 81ε0
- σ = 20 S/m

To find:
Frequency at which the conduction current density is 10 times the displacement current density in magnitude

Solution:
Let's start by using the equations for conduction current density (Jc) and displacement current density (Jd):

Jc = σE
Jd = ε(dE/dt)

Where E is the electric field strength.

We can rewrite Jc in terms of E and Jd by substituting E = Jd/(εω), where ω is the angular frequency:

Jc = σJd/(εω)

Now we can set the magnitudes of Jc and Jd equal to each other:

|Jc| = 10|Jd|

Substituting the expressions for Jc and Jd:

σJd/(εω) = 10ε(dE/dt)

Simplifying:

dE/dt = σωε/10

This is a first-order differential equation, which we can solve by separation of variables:

∫dE/σωε = ∫dt/10

(E/σωε) = t/10 + C

Where C is a constant of integration. We can set C = 0 since E = 0 when t = 0.

Solving for E:

E = σωεt/10

Now we can find the frequency at which |Jc| = 10|Jd| by substituting this expression for E into the expression for Jd and solving for ω:

Jd = εωE = εω(σωεt/10)

Jc = σE = σ(σωεt/10)

|Jc| = 10|Jd|

σ(σωεt/10) = 10εω(σωεt/10)

Simplifying:

ω^2 = 100/81σε

ω = 10/9√(σε)

Substituting the given values:

ω = 10/(9√(20*81ε0*μ0))

ω = 36.02 GHz

Therefore, the frequency at which the conduction current density is 10 times the displacement current density in magnitude is 36 GHz.

Similar GATE Doubts

The Bode plot of a system is shown in figure below the magnitude (in dB) at ω = 10 rad/sec is___________ (Assume initial low frequency is 0.5 rad/se c) Correct answer is between '24,25'. Can you explain this answer?

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