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# Test: Displacement & Conduction Current

## 10 Questions MCQ Test Electromagnetic Fields Theory | Test: Displacement & Conduction Current

Description
This mock test of Test: Displacement & Conduction Current for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam. This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Test: Displacement & Conduction Current (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Displacement & Conduction Current quiz give you a good mix of easy questions and tough questions. Electronics and Communication Engineering (ECE) students definitely take this Test: Displacement & Conduction Current exercise for a better result in the exam. You can find other Test: Displacement & Conduction Current extra questions, long questions & short questions for Electronics and Communication Engineering (ECE) on EduRev as well by searching above.
QUESTION: 1

### Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.

Solution:

Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.

QUESTION: 2

### Calculate the displacement current density when the electric flux density is 20sin 0.5t.

Solution:

Explanation: The displacement current density is given by, Jd = dD/dt.
Jd = d(20sin 0.5t)/dt = 20cos 0.5t (0.5) = 10cos 0.5t.

QUESTION: 3

### Find the magnitude of the displacement current density in air at a frequency of 18GHz in frequency domain. Take electric field E as 4 units.

Solution:

Explanation: Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier transform, Jd = εjωE. The magnitude of
Jd = εωE = ε(2πf)E. On substituting, we get 4 ampere.

QUESTION: 4

Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.

Solution:

Explanation: When Jd = Jc , we get εωE = σE. Thus εo(2∏f) = σ. On substituting conductivity as one and permittivity as 2, we get f = 9GHz.

QUESTION: 5

The ratio of conduction to displacement current density is referred to as

Solution:

Explanation: Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The loss tangent is used to determine if the material is a conductor or dielectric.

QUESTION: 6

If the loss tangent is very less, then the material will be a

Solution:

Explanation: If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor and the material should be a dielectric. Since it is specifically mentioned very less, assuming the conductivity to be zero, the dielectric will be lossless (ideal).

QUESTION: 7

In good conductors, the electric and magnetic fields will be

Solution:

Explanation: The electric and magnetic fields will be out of phase by 45 in good conductors. This is because their intrinsic impedance is given by η = √(ωμ/σ) X (1+j). In polar form we get 45 out of phase.

QUESTION: 8

In free space, which of the following will be zero?

Solution:

Explanation: In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be unity. Since the free space will contain no charges in it, the conductivity will be zero.

QUESTION: 9

If the intrinsic angle is 20, then find the loss tangent.

Solution: