This set of Electromagnetic Theory Interview Questions and Answers for Experienced people focuses on “Displacement and Conduction Current”.
1. Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.
Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.
2. Calculate the displacement current density when the electric flux density is 20sin 0.5t.
a) 10sin 0.5t
b) 10cos 0.5t
c) 20sin 2t
d) 20cos 2t
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.
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.
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.
4. Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.
a) 18 GHz
b) 9 GHz
c) 36 GHz
d) 24 GHz
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.
5. The ratio of conduction to displacement current density is referred to as
a) Attenuation constant
b) Propagation constant
c) Loss tangent
d) Dielectric constant
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.
6. If the loss tangent is very less, then the material will be a
b) Lossless dielectric
c) Lossy dielectric
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).
7. In good conductors, the electric and magnetic fields will be
a) 45 in phase
b) 45 out of phase
c) 90 in phase
d) 90 out of phase
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.
8. In free space, which of the following will be zero?
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.
9. If the intrinsic angle is 20, then find the loss tangent.
a) tan 20
b) tan 40
c) tan 60
d) tan 80
Explanation: The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.
10. The intrinsic impedance of free space is given by
a) 272 ohm
b) 412 ohm
c) 740 ohm
d) 377 ohm
Explanation: The intrinsic impedance is given by η = √(μo/εo) ohm. Here εo = 8.854 x 10-12 and μo = 4π x 10-7.
On substituting the values, we get η = 377 ohm.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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