GATE Electronics and Communication Engineering - Electromagnetic Field Theory
Exam Duration: 45 Mins Total Questions : 30
The equivalent coordinate of
\(\overset { \rightarrow }{ A } =2\overset { \rightarrow }{ a_{ x }-2 } \overset { \rightarrow }{ a_{ y } } +3\overset { \wedge }{ a } _{ z }\)
In cylindrical coordinate system is
- (a)
\(\left( \sqrt { 17 } ,\frac { 3\pi }{ 4 } ,3 \right) \)
- (b)
\(\left( \sqrt { 17 } ,\frac { \pi }{ 4 } ,3 \right) \)
- (c)
\(\left( \sqrt { 8 } ,\frac { \pi }{ 4 } ,3 \right) \)
- (d)
\(\left( \sqrt { 8 } ,\frac { 3\pi }{ 4 } ,3 \right) \)
We know that , in cylindrical coordinate system,
\(\rho =\sqrt { x^{ 2 }+y^{ 2 } } =\sqrt { 2^{ 2 }+2^{ 2 } } =\sqrt { 8 } \)
\(\phi =tan^{ -1 }\left( \frac { y }{ x } \right) =tan^{ -1 }(-1)=\frac { 3\pi }{ 4 } \)
Consider \(\overset { \rightarrow }{ A } =3\overset { \wedge }{ a } _{ x }+4\overset { \wedge }{ a } _{ y }\) and \(\overset { \rightarrow }{ B } =7\overset { \wedge }{ a } _{ x }+\overset { \wedge }{ 2a } _{ y }\) The smaller angle between the two vectors \(\overset { \rightarrow }{ A } \quad \overset { \rightarrow }{ B } \) will be
- (a)
\(39.72^{ 0 }\)
- (b)
\(27.93^{ 0 }\)
- (c)
\(41.9^{ 0 }\)
- (d)
\(19.4^{ 0 }\)
\(\overset { \rightarrow }{ A } X\overset { \rightarrow }{ B } =\left| _{ x }\quad a_{ y }\quad \quad \quad a_{ z }\\ 3\quad 4\quad \quad \quad \quad 0\\ 0\quad \quad 7\quad \quad -2 \right| =-8a_{ x }+6a_{ y }+23a_{ z }\)
\(\left| \overset { \rightarrow }{ A } \right| =\sqrt { 3^{ 2 }+4^{ 2 }+0^{ 2 } } =5\)
\(\left| \overset { \rightarrow }{ B } \right| =\sqrt { 7^{ 2 }+(-2)^{ 2 } } =7.28\)
\(\left| AXB \right| =\sqrt { (-8)^{ 2 }+(6)^{ 2 }+(21)^{ 2 } } \)
\(\left| \overset { \rightarrow }{ A } X\overset { \rightarrow }{ B } \right| =23.25\)
Since \(\left| \overset { \rightarrow }{ A } X\overset { \rightarrow }{ B } \right| =\left| \overset { \rightarrow }{ A } \right| \left| \overset { \rightarrow }{ B } \right| sin0\)
\(\therefore sin\vartheta =\frac { 23.25 }{ 5X7.28 } =0.638\)
\(\therefore \vartheta =39.7^{ 0 }\)
Maxwell's divergence equation for the magnetic field is given by
- (a)
\(\nabla X\overset { \rightarrow }{ B } =0\)
- (b)
\(\nabla .\overset { \rightarrow }{ B } =0\)
- (c)
\(\nabla .\overset { \rightarrow }{ B } =p\)
- (d)
\(\nabla X\overset { \rightarrow }{ B } =p\)
For non-existence of isolated magnetic change Maxwell's equation is
\(\nabla .\overset { \rightarrow }{ B } =0\)
On the boundary the reflected energy is
- (a)
6.25%
- (b)
12.5%
- (c)
25%
- (d)
50%
\(\epsilon _{ n }=\left( \frac { 2\pi C }{ \lambda _{ 1 }\omega } \right) \epsilon _{ n2 }=\left( \frac { 2\pi C }{ \lambda _{ 2 }\omega } \right) ^{ 2 }\Rightarrow \frac { \epsilon _{ n } }{ \epsilon _{ r2 } } \)
\(=\left( \frac { \lambda _{ 2 } }{ \lambda _{ 1 } } \right) \)
\(=\frac { n_{ 2- }n_{ 1 } }{ n_{ 2 }+n_{ 1 } } -\frac { \frac { n_{ 0 } }{ \sqrt { \epsilon _{ n } } } -\frac { n_{ 0 } }{ \sqrt { \epsilon _{ 1 } } } }{ \frac { n_{ 0 } }{ \sqrt { \epsilon _{ n } } } -\frac { n_{ 0 } }{ \sqrt { \epsilon _{ 2 } } } } =\frac { \frac { \epsilon _{ n } }{ \epsilon _{ r2 } } -1 }{ \sqrt { \frac { \epsilon _{ n } }{ \epsilon _{ r2 } } +1 } } \)
Th fraction of the incident energy that is reflected is
\(=\frac { 1 }{ 16 } =6.25%\)
\(\triangledown X(\triangledown .\overset { \rightarrow }{ A } )\quad \) is
- (a)
always a scalar
- (b)
always a vector
- (c)
meaningless
- (d)
can be either a vector or a scalar
\(\triangledown .\overset { \rightarrow }{ A } \) is always a scalar For taking cross product we require 2 vectors Hence \(\triangledown X(\triangledown .\overset { \rightarrow }{ A } )\quad \) will be a meaningless quantity.
Which of the following field equations indicate that the free magnetic charges do not exist ?
- (a)
\(\overset { \rightarrow }{ H } =\int { \frac { IdIXr }{ 4\pi r^{ 2 } } } \)
- (b)
\(\overset { \rightarrow }{ H } =\frac { 1 }{ \mu } (\triangledown X\overset { \rightarrow }{ A } )\)
- (c)
\(\triangledown .\overset { \rightarrow }{ H } =0\)
- (d)
\(\triangledown .\overset { \rightarrow }{ H } =j\)
Gauss's law for magnetostatic field is
Indicates no free magnetic charges exist.
The work done in carrying a charge through an equipotential surface
- (a)
is zero
- (b)
depends on the charge Q
- (c)
is infinity
- (d)
depends on the distance
Since, the potential difference between any two points on the equipotential surface is zero work done is also zero
A rectangular coil of 200 turns has dimensions 0.3X0.15 m with a current of 5 A , placed in an uniform field of 0.2 T Then magnetic moment \(\overset { \rightarrow }{ m } \) is
- (a)
4.5 A-\(m^{ 2 }\)
- (b)
45 A-\(m^{ 2 }\)
- (c)
0.45 A-\(m^{ 2 }\)
- (d)
0.225 A-\(m^{ 2 }\)
The magnetic moment \(\overset { \rightarrow }{ m } \) for a current loop is given by \(m=1A\overset { \rightarrow }{ a } _{ n }\)
\(\overset { \rightarrow }{ m } =5X0.3X0.15=0.225A-m^{ 2 }\)
or 200 turns of coil \(\overset { \rightarrow }{ m } =NIA\overset { \rightarrow }{ a } _{ n }\)
\(\overset { \rightarrow }{ m } =200X0.225=45\quad A-m^{ 2 }\)
For a short circuited coaxical transmission line Characteristic impedence \(Z_{ 0 }=35+j49\Omega \) Propagation constant \(\gamma =1.4+j5m\) Length of line =0.4 m
The input impedence of short circuited line is
- (a)
82+j39\(\Omega \)
- (b)
41+j78\(\Omega \)
- (c)
68+j46\(\Omega \)
- (d)
34+j23\(\Omega \)
\(Z_{ in }=Z_{ ac }=Z_{ 0 }tanh\quad \gamma l=Z_{ 0 }\frac { sinh\gamma l }{ cosh\gamma l } \)
\(\gamma l=(1.4+j5)(0.4)=0.56+j2\)
\(sinh\gamma l=-0.483+j0.536\)
\(Z_{ in }=\frac { (35+j49)(-0.245+j1.055) }{ (-0.483+j0.536) } =82+j39\Omega \)
Some unknown material has a conductivity \(10^{ 6 }\) mho/m and a permeability of \(4\pi X10^{ 7 }\) H/m The skin depth for the material at 1 GHz is
- (a)
15.9 \(\mu m\)
- (b)
25.9 \(\mu m\)
- (c)
20.9 \(\mu m\)
- (d)
30.9 \(\mu m\)
\(\delta =\frac { 1 }{ \sqrt { \pi f\mu \sigma } } =\frac { 1 }{ \sqrt { \pi X1x10^{ 9 }X4\pi x10^{ -7 }X10^{ 6 } } } \)
\(\delta =15.9\mu m\)
The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the \(TE_{ 10 }\) mode is
- (a)
equal to its group velocity
- (b)
less than the velocity of light in free space
- (c)
equal to the velocity of light in free space
- (d)
greater then the velocity of light in free space
Phase velocity\((\overset { \rightarrow }{ v } _{ p })=\frac { C }{ \sqrt { 1-(\lambda _{ 0 }/\lambda _{ e })^{ 2 } } } \)
As \(\sqrt { \frac { \lambda _{ 0 } }{ \lambda _{ c } } } <1\)
Hence \(\overset { \rightarrow }{ v } _{ p }\)>c (velocity of light)
The propagation constant of a lossy transmission line \(\gamma =1+j2m^{ -1 }\) and characteristic impedence is 20 \(\Omega \) at frequency 1 Mrad/s
A line comprised of two copper wires of diameter 1.2 mm that have 3.2 mm centre to centre spacing If the wires are seperated by a dielectric material with \(\varepsilon _{ r }=35\) the value of characteristic impedence \(Z_{ 0 }\) is
- (a)
96 \(\Omega \)
- (b)
150 \(\Omega \)
- (c)
75 \(\Omega \)
- (d)
105 \(\Omega \)
2a=1.2 mm,d=3.2 mm
\(L=\frac { \mu }{ \pi } cosh^{ -1 }\left( \frac { d }{ 2a } \right) \)
=\(4X10^{ 7 }Cosh^{ -1 }\left( \frac { 3.2 }{ 1.2 } \right) =0.6\mu H/m\)
C=\(C=\frac { \pi \varepsilon }{ cosh^{ -1 }\left( \frac { d }{ 2a } \right) } =\frac { 10^{ -9 }X3.5 }{ cosh^{ -1 }\left( \frac { 3.2 }{ 1.2 } \right) } =59.4\quad pF/m\)
\(Z_{ 0 }=\sqrt { \frac { L }{ C } } =\sqrt { \frac { 0.66X10^{ -6 } }{ 59.4X10^{ -12 } } } =105\Omega \)
For a wave propagating in air filled rectangular waveguide
- (a)
Guided wavelength is never less than the free space wavelength
- (b)
Wave impedance is never less than the free space impedence
- (c)
Phase velocity is never less than the free space velocity
- (d)
TEM mode is possible if the dimensions of the waveguide are properly chosen
1.Guided wavelength is never less than the free space wavelength
2.phase velocity is never less than the free space velocity
In the design of a single mode step index optical fibre close to upper cut-off the single mode operations is not preserved if
- (a)
radius as well as operating wavelengths are halved
- (b)
radius as well as and operating wavelength are doubled
- (c)
radius is halved and operating wavelength is doubled
- (d)
radius is doubled and operating wavelength is halved
The number of modes supported by a fibre is determined by cut-off parameter called V-number Below that value the mode will cut-off
\(V=\frac { \pi d }{ \lambda _{ 0 } } (NA)\)
If radius is doubled and operating wavelength is halved the V number becomes \(\frac { 1 }{ 4 } \) times of design V-number and single mode operation is not preserved.
Which one of the following does represent the electric field lines for the \(TE_{ 02 }\) mode in the cross section of a hollow rectangular metallic waveguide?
- (a)
- (b)
- (c)
- (d)
The electric field line for the \(TE_{ 02 }\) mode in the rectangular metallic waveguide will be
A 60 \(\Omega \) coaxial cable feeds a 75+j25\(\Omega \) dipole antenna.The voltage reflection coefficient and\(\Gamma \) standing wave ratio S are respectively.
- (a)
0.212 \(<48.55^{ 0 },1.538\)
- (b)
0.486 \(<68.4^{ 0 },2.628\)
- (c)
0.486 \(<68.4^{ 0 },2.628\)
- (d)
0.212 \(<68.4^{ 0 },1.538\)
\(\Gamma =\frac { Z_{ L }-{ Z }_{ 0 } }{ Z_{ L }+{ Z }_{ 0 } } =\frac { 75+j25-60 }{ 75+j25+60 } =0.212<48.55^{ 0 }\)
\(S=\frac { 1+\left| \Gamma \right| }{ 1-\left| \Gamma \right| } =\frac { 1+0.212 }{ 1-0.212 } =1.538\)
The unit vector directed from point A(5,-1,0) towards point B(3,0,2) is
- (a)
\(-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ X }+\frac { 1 }{ 3 } \overset { \wedge }{ u } _{ y }+-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ z }\)
- (b)
\(-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ X }-\frac { 1 }{ 3 } \overset { \wedge }{ u } _{ y }+-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ z }\)
- (c)
\(-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ X }+\frac { 1 }{ 3 } \overset { \wedge }{ u } _{ y }+\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ z }\)
- (d)
\(\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ X }-\frac { 1 }{ 3 } \overset { \wedge }{ u } _{ y }-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ z }\)
\(\overset { \rightarrow }{ R } _{ AB }=\overset { \rightarrow }{ R } _{ B }-\overset { \rightarrow }{ R } _{ A }\)
\(=(3\overset { \wedge }{ u } _{ X }+0\overset { \wedge }{ u } _{ y }+2\overset { \wedge }{ u } _{ z })-(5\overset { \wedge }{ u } _{ X }-\overset { \wedge }{ u } _{ y }+0\overset { \wedge }{ u } _{ Z })\)
\(\quad \quad \quad =-2\overset { \wedge }{ u } _{ X }+\overset { \wedge }{ u } _{ y }+2\overset { \wedge }{ u } _{ z\\ }\)
\(\left| \overset { \rightarrow }{ R } _{ AB } \right| =\sqrt { 2^{ 2 }+1+2^{ 2 } } =3\)
\(\overset { \wedge }{ u } _{ R }=-\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ x }+\frac { 1 }{ 3 } \overset { \wedge }{ u } _{ y }+\frac { 2 }{ 3 } \overset { \wedge }{ u } _{ z }\)
The vector component of \(\overset { \rightarrow }{ F } =y\overset { \wedge }{ u_{ x } } +z\overset { \wedge }{ u_{ y } } +\overset { \wedge }{ u_{ z } } \) that is perpendicular to \(\overset { \rightarrow }{ G } =4\overset { \rightarrow }{ u } _{ x }+4\overset { \rightarrow }{ u } _{ y }-2\overset { \rightarrow }{ u } _{ z }\) is
- (a)
\(-3\overset { \rightarrow }{ u } _{ x }-2\overset { \rightarrow }{ u } _{ y }-10\overset { \rightarrow }{ u } _{ z }\)
- (b)
\(-\overset { \rightarrow }{ u } _{ x }-2\overset { \rightarrow }{ u } _{ y }\)
- (c)
\(\overset { \rightarrow }{ u } _{ x }+2\overset { \rightarrow }{ u } _{ z }\)
- (d)
\(\overset { \rightarrow }{ 2u } _{ x }+2\overset { \rightarrow }{ u } _{ y }\)
The vector component of \(\overset { \rightarrow }{ F } \) perpendicular to \(\overset { \rightarrow }{ G } \)
\(=\overset { \rightarrow }{ F- } \frac { \overset { \rightarrow }{ F. } \overset { \rightarrow }{ G. } }{ G^{ 2 } } \overset { \rightarrow }{ G } \)
\(=(3,2,1)-\frac { (3,2,1)(4,4,-2) }{ 4^{ 2 }+4^{ 2 }+2^{ 2 } } (4,4,-2)\)
\(=(3,2,1)-(2,2-1)=(1,0,2)=\overset { \rightarrow }{ u } _{ x }+2\overset { \rightarrow }{ u } _{ z }\)
The electric field component of a time harmonic plane EM were travelling in a non-magnetic lossless dielectric medium has an amplitude 1 V/m .if the relative permittivity of the medium is 4, the magnitude of the time average power density vector (in \(W/m^{ 2 })\) is
- (a)
\(\frac { 1 }{ 30\pi } \)
- (b)
\(\frac { 1 }{ 60\pi } \)
- (c)
\(\frac { 1 }{ 120\pi } \)
- (d)
\(\frac { 1 }{ 240\pi } \)
From Poynting theorem,
\(\overset { \rightarrow }{ P } _{ av }=\frac { 1 }{ 2 } (\overset { \rightarrow }{ E } X\overset { \rightarrow }{ H } *)\)
\(\left| \overset { \rightarrow }{ P } _{ av } \right| =\frac { 1 }{ 2 } EH(W/m^{ 2 })\)
\(H=E/n\quad and\quad n=\sqrt { \mu /\epsilon } \\ \)
\(\left| \overset { \rightarrow }{ P_{ av } } \right| =\frac { 1 }{ 2 } \frac { E^{ 2 } }{ \eta } =\frac { E^{ 2 } }{ 2\sqrt { \frac { \mu ,\mu _{ 0 } }{ \epsilon _{ r },\epsilon _{ 0 } } } } \)
Given \(\epsilon _{ r }=4\)
non-magnetic field \(\mu _{ r }=1\)
\(\left| \overset { \rightarrow }{ P_{ av } } \right| =\frac { E^{ 2 } }{ 2\sqrt { \frac { \mu _{ 0 } }{ \epsilon _{ 0 } } X\frac { 1 }{ 4 } } } \)
\(\frac { 1 }{ 120\pi } (W/m^{ 2 })\)
The surface r=4 and 8, \(\theta =60^{ 0 },90^{ 0 },\phi =40^{ 0 }\) and \(100^{ 0 }\) identify a closed surface The enclosed volume is
- (a)
70
- (b)
80.85
- (c)
78.19
- (d)
76.77
Volume=\(\int _{ 40 }^{ 100 }{ \int _{ 60 }^{ 90 }{ \int _{ -4 }^{ 8 }{ r^{ 2 }sin\theta drd\theta d\phi } } } \)
\(=\frac { 448 }{ 3 } (0.5)(60^{ 0 })\)
\(=\frac { 448 }{ 3 } (0.5)\left( \frac { \pi }{ 3 } \right) =78.19\)
Refractive index of glass is 1.5 Find the wavelength of a beam of light with a frequency of \(10^{ 14 }\) Hz in glass.Assume velocity of light is \(3X10^{ 8 }m/s\) in vaccum
- (a)
3 \(\mu m\)
- (b)
3 mm
- (c)
2 \(\mu m\)
- (d)
1 \(\mu m\)
\({ V }_{ G }=\)Velocity of light in glass
=\(-\frac { Velocity\quad in\quad vaccum }{ Refractive\quad Index } =\frac { c }{ \mu } \)
\(\lambda =\frac { v_{ g } }{ f } =\frac { c }{ \mu f } =\frac { 3X10^{ 8 } }{ 1.5X10^{ 14 } } =2X10^{ -6 }m=2\mu m\)
The electric field of a uniform plane wave in free space is given by \(\overset { \rightarrow }{ E_{ s }= } \quad 12\pi (\mu _{ y }+j\mu _{ z })e^{ -ji5x }\) The magnetic field phasor is
- (a)
\(\frac { 12 }{ \eta _{ 0 } } (-\mu _{ z }+j\mu _{ y })e^{ -\beta 5x }\)
- (b)
\(\frac { 12 }{ \eta _{ 0 } } (-\mu _{ z }-j\mu _{ y })e^{ -\beta 5x }\)
- (c)
\(\frac { 12 }{ \eta _{ 0 } } (\mu _{ z }-j\mu _{ y })e^{ -\beta 5x }\)
- (d)
\(\frac { 12 }{ \eta _{ 0 } } (\mu _{ z }+j\mu _{ y })e^{ -\beta 5x }\)
Since Poynting vector is in the positive X-direction Therefore \(\mu _{ E }X\mu _{ H }=\mu _{ X }\)
\(\mu _{ E }=\mu _{ y },u_{ y }X\mu _{ H }=\mu _{ x }\)
\(\Rightarrow \mu _{ H }=\mu _{ Z }\)
For \(\mu _{ E }=\mu _{ y },u_{ y }\mu _{ H }=\mu _{ x }\)
\(\mu _{ H }=-\mu \)
\(\overset { \rightarrow }{ H } =\frac { 12 }{ \eta _{ 0 } } (\mu _{ z }-j\mu _{ y })e^{ -j15x }\)
If a vector field \(\overset { \rightarrow }{ V } \) is related to another vector field \(\overset { \rightarrow }{ A } \) through \(\overset { \rightarrow }{ V } =\triangledown X\overset { \rightarrow }{ A } \) Which of the following is true?
- (a)
\(\oint { \overset { \rightarrow }{ V } } d\overset { \rightarrow }{ I } =\int { _{ sc }\int { \overset { \rightarrow }{ A } } } .d\overset { \rightarrow }{ s } \)
- (b)
\(\oint { \overset { \rightarrow }{ A } } d\overset { \rightarrow }{ I } =\int { _{ sc }\int { \overset { \rightarrow }{ V } } } .d\overset { \rightarrow }{ s } \)
- (c)
\(\oint { \triangledown X\overset { \rightarrow }{ V } } .d\overset { \rightarrow }{ I } =\int { _{ sc }\int { \triangledown } } \overset { \rightarrow }{ A } .d\overset { \rightarrow }{ s } \)
- (d)
\(\oint { \triangledown X\overset { \rightarrow }{ A } } .d\overset { \rightarrow }{ I } =\int { _{ sc }\int { \triangledown } } \overset { \rightarrow }{ V } .d\overset { \rightarrow }{ s } \)
Given, \(\overset { \rightarrow }{ V } =\triangledown X\overset { \rightarrow }{ A } \)
Taking surface integral
\(\int { _{ sc }\int { \overset { \rightarrow }{ V } } } .d\overset { \rightarrow }{ s } =\int { \int { (\triangledown X\overset { \rightarrow }{ A } } } )d\overset { \rightarrow }{ s } \)
By stokers theorem
\(\int { _{ sc }\int { \overset { \rightarrow }{ V } } } .d\overset { \rightarrow }{ s= } \oint { _{ c }\overset { \rightarrow }{ A } } .d\overset { \rightarrow }{ L } \\ \int { _{ dc }\int { (\triangledown XA). } } d\overset { \rightarrow }{ s= } d\overset { \rightarrow }{ L } \)
A mast antenna consisting of a 50 m long vertical conductor operates over a perfectly conducting ground plane.It is base fed at a frequency of 600 KHz .The radiation resistance of the antenna in ohm is
- (a)
\(\frac { 2\pi ^{ 2 } }{ 5 } \)
- (b)
\(\frac { \pi ^{ 2 } }{ 5 } \)
- (c)
\(\frac { 4\pi ^{ 2 } }{ 5 } \)
- (d)
\(20\pi ^{ 2 }\)
The radiation resistance is given by
\(R_{ RAD }=80\pi ^{ 2 }\left( \frac { 1 }{ \lambda } \right) ^{ 2 }\)
\(=\frac { 80\pi ^{ 2 }I^{ 2 }j^{ 2 } }{ c^{ 2 }\\ } \\ =\frac { 80X\pi ^{ 2 }X50^{ 2 }X600^{ 2 }X10^{ 6 } }{ 3^{ 2 }X10^{ 16 } } \\ =\frac { 8X25X4X10^{ 13 } }{ 9X10^{ 16 } } \pi ^{ 2 }=\frac { 4 }{ 5 } \pi ^{ 2 }\)
The cathode of a planner vaccum tube is at z=0 Let \(\overset { \rightarrow }{ E } =-4X10^{ 6 }\overset { \wedge }{ u } _{ z }\) V/m for Z>0 an electron is emitted from the cathode with zero initial velocity at t=0 The velocity of electron time t is
- (a)
\(4.0X10^{ 6 }tu_{ z }\quad m/s\)
- (b)
\(4.39X10^{ 16 }tu_{ z }\quad m/s\)
- (c)
\(4.46X10^{ 17 }tu_{ z }\quad m/s\)
- (d)
\(7.0X10^{ 17 }tu_{ z }\quad m/s\)
\(\overset { \rightarrow }{ F } =ma=q\overset { \rightarrow }{ E } \\ \overset { \rightarrow }{ a } =\frac { q\overset { \rightarrow }{ E } }{ m } =\frac { (-1.062X10^{ 9 }) }{ 9.11X10^{ -3 } } \)
At 20 GHz, the gain of a parabolic dish antenna of 1 m diametere and 70% efficiency is
- (a)
15 dB
- (b)
25 dB
- (c)
35 dB
- (d)
45 dB
Gain of parabolic dish antenna
\(G=\eta \pi ^{ 2 }\left( \frac { D }{ \lambda } \right) ^{ 2 }\)
\(\lambda =\frac { c }{ f } =\frac { 3X10^{ 8 } }{ 20X10^{ 9 } } =0.015\quad m\)
\(\eta =efficiency=\frac { 70 }{ 100 } =0.7\)
\(G=0.7X(3.14)^{ 2 }\left( \frac { 1 }{ 0.015 } \right) ^{ 2 }\)
=30674.3
The potential function for the region between two concentric right circular cylinders withy r=-1 and 20 and V=0 and 150 V ,respectively is
- (a)
V=50.1 \(log_{ e }(r)\)
- (b)
V=50.1 \(log_{ e }(r)\) +34.59
- (c)
V=150 V
- (d)
\(V=\frac { 50.1 }{ log_{ e }r } V\)
The potential is constantly w.r.t \(\Phi \quad \) and z hence the Laplace\s equation reduce to
\(\frac { 1 }{ r } \frac { d }{ dr } \left( \frac { rdV }{ dr } \right) =0\)
Integrating we get
\(\int { \frac { 1 }{ dr } \left( r\frac { dV }{ dr } \right) } =0\)
\(\frac { rdV }{ dr } =A\)
Again integrating \(\int { dV=\int { \frac { A }{ r } dr } \frac { 1 }{ dr } \left( r\frac { dV }{ dr } \right) } \)
Applying boundary conditions and solving with different values of r and V we get
A=5.0.1 and B=0
V=50.1 In (r)
The phase velocity for the TE mode in an air-filled rectangular waveguide is (C is the velocity of plane waves in free space)
- (a)
less than c
- (b)
equal to C
- (c)
greater than c
- (d)
none of these
As given that the arrangement is end-fire array
\(\Psi =0\)
\(d=\frac { \pi }{ 4 } \)
\(\theta =60^{ 0 }\)
\(\Psi =\beta dcos\theta +\eth \)
\(0=\frac { 2\pi }{ \lambda } .\frac { \lambda }{ 4 } cos60^{ 0 }+\eth \)
\(\eth =-\frac { \pi }{ 4 } \)
A rectangular metal waveguide filled with a dielectric material of relative permitivity has the inside dimensions 3.0cmX1.2cm The cut-off frequency for the dominant mode is
- (a)
2.5 GHz
- (b)
5.0 GHz
- (c)
10.0 GHz
- (d)
12.5 GHz
\(\mu =\frac { c }{ \sqrt { \epsilon _{ 0 } } } =\frac { 3\times 10^{ 8 } }{ 2 } =1.5\times 10^{ 8 }\)
In rectangular waveguide the dominant mode is TE and
\(f_{ 0c }=\frac { V }{ 2 } \sqrt { \frac { m }{ a } +\frac { n }{ b } } \)
\(=\frac { 1.5X10^{ 8 } }{ 2 } \sqrt { \frac { 1 }{ 0.03 } ^{ 2 }+\frac { 0 }{ b } ^{ 2 } } \)
\(=\frac { 1.5\times 10^{ 8 } }{ 0.06 } =2.5\quad GHz\)
A rectangular waveguide has dimensions 1cmX0.5cm Its cut off frequency is
- (a)
58 GHz
- (b)
10 GHz
- (c)
15 GHz
- (d)
12 GHz
Cut-off frequency
\(f_{ 0c }=\frac { V }{ 2 } \sqrt { \frac { m }{ a } +\frac { n }{ b } } \)
For rectangular waveguide dominant mode is \(TE_{ 01 }\)
Thus, \(f_{ c }=\frac { V_{ p } }{ 2a } \)
\(\frac { 3\times 10^{ 8 } }{ 2\times 10^{ -2 } } =15\times 10^{ 9 }\)
=15 GHz