Antenna Theory Questions and Answers Part-10

1. The minus sign in the Faradays law of induction is given by ______
a) Lens Law
b) Gauss law
c) Amperes Law
d) Gauss law

Answer: a
Explanation: Lens law gives only the reason for the negative sign in the Faradays law of induction
emf=\(-\frac{\partial \phi}{\partial t}\)
The minus sign indicates the direction of induced current.

2. Which of the following Maxwell equation is obtained from Amperes law?
a) ∇×H=J
b) emf=\(-\frac{\partial \phi}{\partial t}\)
c) ∇×E=\(\frac{-\partial B}{\partial t}\)
d) ∇×D=ρv

Answer: a
Explanation:
Faradays Law                : ∇×E=\(-\frac{\partial B}{\partial t}\)
Amperes Law                : ∇×H=J+\(\frac{\partial D}{\partial t}\)
Gauss Law for electric field     : D=ρv
Gauss law for magnetic field  : ∇.B=0

3. Gauss for the Magnetic Field is given by ______
a) ∇.B=0
b) ∇×B=ρv
c) ∇×B=0
d) ∇.B=ρv

Answer: a
Explanation: Gauss law for magnetic field states that the net flux out of any closed surface is zero.
∇.B=0
This Maxwell equation is one of the equation used to determine the boundary conditions.

4. Gauss for the Electric Field is given by ______
a) ∇.D=0
b) ∇×D=ρv
c) ∇×D=0
d) ∇.D=ρv

Answer: d
Explanation: Gauss’s law for electric field states that the net flux of an electricfield in a closed surface is directly proportional to the enclosed electric charge. ∇.D=ρv
This Maxwell equation is one of the equation used to determine the boundary conditions.

5. Which of the Following Maxwell equation is for nonexistence of isolated magnetic charge?
a) ∇×E=-\(-\frac{\partial B}{\partial t}\)
b) ∇×H=J
c) ∇.D=ρv
d) ∇.B=0

Answer: d
Explanation: Gauss law for magnetic field states that the net flux out of any closed surface is zero.
∇.B=0
This is satisfied only when two different poles of magnet exist. So this Maxwell equation proves for the nonexistence of the isolated magnetic charge.
Faradays Law : ∇×E=\(-\frac{\partial B}{\partial t}\)
Amperes Law : ∇×H=J
Gauss Law for electric field : ∇.D=ρv

6. In which of the following Integral form of Maxwell equations, the surface is closed?
a) Amperes law
b) Gauss Law
c) Faradays Law
d) Both Amperes and Faraday law

Answer: b
Explanation: The surface integral is closed for the Gauss laws of magnetic and electric fields. It is open for the amperes and Faradays law.
Maxwell Equations:
Gauss law electric field : \(\oint_sD.ds =\int_v\rho_v dv \)
Gauss law magnetic field : \(\oint_sB.ds =0\)
Faradays law : \(\int_cE.dl =-\int_s\frac{\partial B}{\partial t}.dS\)
Amperes law : \(\int_cH.dl =\int_s(\frac{\partial D}{\partial t} + J).dS\)

7. Divergence of Magnetic field is ______
a) volume charge density ρv
b) zero
c) infinite
d) dependent on magnetic field vector

Answer: b
Explanation: The Divergence of Magnetic is always zero.It is obtained from the Maxwell equation ∇.B=0 which is derived from the Gauss law of magnetic field.Gauss law for magnetic field states that the net flux out of any closed surface is zero. ∇.D=ρv.

8. Which of the following Maxwell equation is correct for a non-conducting and lossless medium?
a) ∇.D=ρv
b) ∇.D=0
c) ∇×D=ρv
d) ∇×E=0

Answer: b
Explanation: Since it is given non-conducting medium, the charge density ρv=0 and current density J=0. The Maxwell equations are:
Faradays Law                : ∇×E=\(-\frac{\partial B}{\partial t}\)
Amperes Law                : ∇×H=\(\frac{\partial D}{\partial t}\)
Gauss Law for electric field     : ∇.D=0
Gauss law for magnetic field  : ∇.B=0

9. Find skin depth of 5GHz for silver with a conductivity 6.1×107s/m and relative permittivity 1.
a) 0.00091m
b) 0.9113μm
c) 0.319μm
d) 0.1913μm

Answer: b
Explanation: The skin depth is given by \(δ = \sqrt{\frac{1}{\pi fμσ}}\)
Given f=5GHz
Conductivity σ= 6.1×107 s/m
And μr = 1 =>μ=4π ×10-7
⇨ \(δ = \sqrt{\frac{1}{\pi fμσ}}=0.9113\mu m.\)

10. What is the radiation resistance of an antenna if input power to it is 1KW and current in it is 10A having a power loss of 200W?
a) 10Ω
b) 2Ω
c) 12Ω
d) 8Ω

Answer: d
Explanation: Input power Pin=Prad+Ploss
⇨ Prad=Pin-Ploss=1000-200=800W
Now, Power radiated Prad=Irms2Rrad
⇨ Rrad=\(\frac{P_{rad}}{I_{rms}^2} = \frac{800}{10×10}\)=8Ω