In a coaxial transmission line (εr = 1), the electric field intensity is given by: E = 100 cos ( 10^9t - 6z )u_pV/m The displacement current density is

- 0.9 sin ( 10^9t - 6z )u_pA/m^2

This question belongs to Electronics And Communications Engineering Electromagnetic Field Theory

Electromagnetic Field Theory

In a coaxial transmission line (ε_r = 1), the electric field intensity is given by:
$$E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{V/m}}$$
The displacement current density is

A) $$ - \frac{{100}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$ B) $$\frac{{116}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$ C) $$ - \frac{{0.9}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$ D) $$ - \frac{{216}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$

Answer & Solution

Correct Answer: Option C

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In a coaxial transmission line ( r = 1) the electric...

In a coaxial transmission line (εr = 1), the electric field intensity is given by: $$E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{V/m}}$$ The displacement current density is

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Related Questions on Electromagnetic Field Theory

In a coaxial transmission line (ε_r = 1), the electric field intensity is given by:
$$E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{V/m}}$$
The displacement current density is