In a travelling electromagnetic wave, E and H vector fields are .... In an electromagnetic wave, the phase difference between electric and magnetic field vectors.
AE14
ELECTROMAGNETICS AND RADIATION
TYPICAL QUESTIONS & ANSWERS PART – I
OBJECTIVE TYPES QUESTIONS Each Question carries 2 marks. Choose correct or the best alternative in the following: Q.1
Two concentric spherical shells carry equal and opposite uniformly distributed charges over their surfaces as shown in Fig.1. Electric field on the surface of inner shell will be (A) zero. Q (B) . 4 Π ∈0 R 2 Q (C) . 8 Π ∈0 R 2 Q (D) . 16 Π ∈0 R 2 Ans: A
Q.2
The magnetic field intensity (in A
m
and carrying current of 2 A is (A) 8. (C) 3.
) at the centre of a circular coil of diameter 1 metre
(B) 4. (D) 2.
Ans: A Q.3
The polarization of a dielectric material is given by →
→
→
→
→
(B) P = (∈r −1) E .
(A) P =∈r E . →
→
(C) P = E ∈0 (∈r −1) .
(D) P = (∈r −1) ∈0 .
Ans: B Q.4
In a travelling electromagnetic wave, E and H vector fields are (A) perpendicular in space . (B) parallel in space. (C) E is in the direction of wave travel. (D) H is in the direction of wave travel.
Ans: C 1
AE14 Q.5
ELECTROMAGNETICS AND RADIATION For a broad side linear array which of the following is not correct
(A) the maximum radiation occurs perpendicular to the line of the array at φ = 90o . (B) the progressive phase shift (α ) between elements is zero. (C) width of principal lobe is less than that of an end fire array. (D) the maximum radiation occurs along the line of array at φ = 0o . Ans: D Q.6
Two point charges Q1 and Q 2 are located at A and B on a straight line as shown in the Fig.1. The electric field will be zero at a point
(A) between A and B. (C) to the right of B.
(B) to the left of A. (D) perpendicular to AB.
Ans: B Q.7
The ratio of conduction current density to the displacement current density is σ jσ (A) (B) jω ∈ ω∈ σω σ∈ (C) (D) j∈ jω
Where symbols have their usual meaning.
Ans: A Q.8
A wave is incident normally on a good conductor. If the frequency of a plane electromagnetic wave increases four times, the skin depth, will (A) increase by a factor of 2. (B) decrease by a factor of 4. (C) remain the same. (D) decrease by a factor of 2.
Ans: D Q.9
(
)
Electric field of a travelling wave is given by 100 cos 109 t − 4x . The velocity and the wavelength are (A) 3 × 108 m and 4 m respectively. sec (B) 2.5 × 108 m and π m respectively. sec 2 9m (C) 4 × 10 and 8 π m respectively. sec (D) 109 m and 100 m respectively. sec 4
2
AE14
ELECTROMAGNETICS AND RADIATION Ans: B
Q.10
In a dielectric-conductor boundary (interface), the tangential component of electric field is (A) E t (B) 2E t (C) zero (D) infinity
Ans: C Q.11
Which of the following matching is incorrect: r r r ∂D (A) Ampere’s circuital law → ∇ × H = J + . ∂ t r r ∂D (B) Displacement current density → J = . ∂t
(C) Poisson’s equation → ∇ 2 V = 0 . r ∂ρ (D) Continuity equation → ∇ ⋅ J = − . ∂t Ans: C Q.12
For a transmission line terminated in its characteristic impedance, which of the following statement is incorrect: (A) It is a smooth line. (B) The energy distribution between magnetic and electric field is not equal. (C) Standing wave does not exist. (D) Efficiency of transmission of power is maximum.
Ans: B Q.13
For a line of characteristic impedance, Z O terminated in a load, Z R such that Z R > Z O , the Voltage Standing Wave Ratio (VSWR) is given by Z . (A) R (B) Z O . ZO Z (C) Z R . (D) O . ZR Ans: A
Q.14
The lower cut-off frequency of a rectangular wave guide with inside dimensions (3 × 4.5 cm) operating at 10 GHz is (A) 10 GHz. (B) 9 GHz. 10 10 (C) GHz. (D) GHz. 9 3 Ans: D
Q.15
The directive gain cannot be stated as (A) the ratio of the radiation intensity in that direction to the average radiated power. (B) the function of angles. 3
AE14
ELECTROMAGNETICS AND RADIATION (C) the directivity of an antenna when directive gain is maximum. (D) independent of angles. Ans: D
Q.16
Electric field intensity due to line charge of infinite length is. ρL ρL (A) . (B) . 2 πεR 4 πεR 2ρ L 2ρL (C) . (D) . πεR 4πεR
Ans: A Q.17
With respect to equipotential surface pick the odd one out. (A) Potential is same every where (B) Work done in moving charge from one point to another is zero (C) Potential is different every where (D) No current flows on this surface
Ans: C Q.18
Energy stored in a magnetic field is
1 (A) W= 2
H2 (B) W = µ . 2
µH . 4
µH 2 (C) W= . 2
(D) W =
µ H . 2
Ans: C Q.19
The intrinsic impedance of free space is (A) 75 ohm. (B) 73 ohm. (C) 120 π ohm. (D) 377ohm.
Ans: D Q.20
During night which layer does not exist? (A) D layer (B) F1 layer (D) E layer (C) F2 layer
Ans: A Q.21
The characteristic impedance is given by Zoc Zsc (A) Z0 = (B) Zsc Zoc (C) Zsc Zoc (D) (Zsc ⋅ Zoc )
Ans: C 4
AE14 Q.22
ELECTROMAGNETICS AND RADIATION Transverse electric wave traveling in z- direction satisfies (A) Ez = 0; Hz = 0 (B) Ez = 0; Hz ≠ 0 (D) Ez ≠ 0; Hz ≠ 0. (C) Ez ≠ 0; Hz = 0
Ans: B Q.23
Radiation resistance of a λ /2 dipole is (A) 73 Ω . (B) 75 ohm (C) 120 π ohm (D) 377 ohm
Ans: A Q.24
Poisson’s equation in CGS Gaussian system is ρ (A) ∇ 2 V = − (B) ∇ 2 V = −4πσ ∈o
(D) ∇ 2 V = 0
(C) ∇ 2 V = −4πρ Ans: C Q.25
The electric potential due to linear quadrapole varies inversely with
(A) r
(B) r 2
(C) r 3
(D) r 4
Ans: C Q.26
In an electromagnetic wave, the phase difference between electric and magnetic field vectors r r E and B is (A) zero (B) π 2 (D) π 4 (C) π
Ans: A
(
)
Q.27
If the electrostatic potential is given by φ = φ o x 2 + y 2 + z 2 where φo is constant, then the charge density giving rise to the above potential would be (A) zero (B) −6φo ∈o ∈ φo (C) −2φo ∈o (D) − ∈o Ans: B
Q.28
Find the amplitude of the electric field in the parallel beam of light of intensity 2.0 W m 2 . (A) 28.8 N c (B) 15.6 N c (C) 38.8 N c (D) 26.6 N c
Ans: C 5
AE14 Q.29
ELECTROMAGNETICS AND RADIATION The material is described by the following electrical parameters at a frequency of 10 GHz, σ σ = 10 6 mho / m, µ = µ o and = 10 , the material at this frequency is considered to be σc (A) a good conductor (B) neither a good conductor nor a good dielectric (C) a good dielectric (D) a good magnetic material
Ans: A Q.30
Consider a transmission line of characteristic impedance 50 ohms and the line is terminated at one end by +j50 ohms, the VSWR produced in the transmission line will be (A) +1 (B) zero (C) infinity (D) -1
Ans: C Q.31
A very small thin wire of length
(A) 0 Ω (C) 7.9 Ω
λ has a radiation resistance of 100 (B) 0.08Ω (D) 790Ω
Ans: B Q.32
Which one of the following conditions will not gurantee a distortionless transmission line (A) R = 0 = G (B) RC = LG (C) very low frequency range (R>> ω L, G >> ω C) (D) very high frequency range (R ω L, G >> ω C (D) R θc
Ans: D Q.65
Poisson’s equation is −ρ (A) ∇ 2 V = ∈
(B) ∇ 2 V = −4πσ (D) ∇ 2 V = 0
(C) ∇ 2 V = −4πρ Ans: A Q.66
The radio wave is incident on layer of ionosphere at an angle of 30o with the vertical. If the critical frequency is 1.2 MHz, the maximum usable frequency (MUF) is (A) 1.2 MHz (B) 2.4 MHz (C) 0.6 MHz (D) 1.386 MHz
Ans: D Q.67
A transmission line with a characteristic impedance Z1 is connected to a transmission line with characteristic impedance Z 2 . If the system is being driven by a generator connected to the first line, then the overall transmission coefficient will be 2 Z1 Z1 (A) (B) Z1 + Z 2 Z1 + Z 2 2Z 2 Z2 (C) (D) Z1 + Z 2 Z1 + Z 2 Ans: A
Q.68
A rectangular waveguide has dimension 1.0 cm × 0.5 cm , its cutoff frequency for the dominant mode is
11
AE14
ELECTROMAGNETICS AND RADIATION (A) 5 GHz (C) 15 GHz
(B) 10 GHz (D) 20 GHz
Ans: C Q.69
Poynting vector gives (A) rate of energy flow. (B) direction of polarization. (C) intensity of electric field. (D) intensity of magnetic field.
Ans:A
12
AE14
ELECTROMAGNETICS AND RADIATION
PART – II
NUMERICALS Q.1
Using Gauss’s theorem, show that a symmetrical spherical charge distribution is equivalent to a concentrated point charge at the centre of the sphere as far as external fields are concerned. (6)
Ans:Refer section 2.21 & 2.22 Pg No 109 of “Electromagnetic Field and Waves” by K D Prasad. Q.2
A spherical volume of radius R has a volume charge density given by ρ = Kr , where r is the →
radial distance and K is a constant. Develop expressions for E and V and sketch their variation with respect to r (0 ≤ r ≤ ∞ ) . (8)
Ans:For r < R : Gauss’s law ⇒ ∈o ∫ Eds = charge enclosed by sphere of radius r. Or ∈o E r 4πr 2 = ∫ ρdv = ∫ kr.4πr 2 dr = 4πk ∫ r 3dr = πkr 4 Or E r =
kr 2 i r for r < R 4 ∈o
At r = R, E R =
kR 2 i r max value at r = R 4 ∈o R
For r ≥ R ; ∈0 E r 4πr 2 = ∫ ρdv 0 R
= 4πk ∫ r 3dr = πKR 4 0 4
Or E r =
Q.3
KR ir 4 ∈0 r 2
for r ≥ R .
Give an example in which the current in a wire enclosed by a closed path is not a uniquely defined. Is it correct to apply Ampere’s circuital law for the static case in such a situation? Explain. (6)
Ans:Refer section 4.2, Page No 306 of “Electromagnetic Field and Waves” by K D Prasad. →
Q.4
The electric field E in free space is given as E = E m cos (ωt − βz ) ˆi x .
13
AE14
ELECTROMAGNETICS AND RADIATION → →
→
(8)
Determine D , B and H . Sketch E and H at t =0.
Ans: D =∈0 E =∈0 E m cos(ωt − βz )i x
ix iy iz ∂B ∂ ∂ ∂ ∇× E = − = ∂t ∂x ∂y ∂z 0 Em cos(ωt − β z ) 0 ∂B ∂ Or − = [Em cos(ωt − β z )]i y ∂t ∂z = + β Em sin (ωt − β z )i y Integrating, in view of its being a static field, the constant of integration is treated as zero. βE m −B=− cos(ωt − β z )i y
ω β Em cos(ωt − β z )i y Or B = ω B β Em H= = cos(ωt − β z )i y . µ 0 ωµ 0
Q.5
What is the physical interpretation of Gauss’s law for the magnetic field? How gauss’ law for the magnetic field in differential form can be derived from it integral form? (3 + 5)
Ans:Refer section 8.5, Page No 251 of Engineering Electromagnetics by William H Hagt, 6th edition. Q.6
d 2V
= 0 . The boundary conditions dx 2 are V = 9 at x = 1 and V = 0 at x = 10. Find the potential and also show the variation of V (6) with respect to x.
The one-dimensional Laplace’s equation is given as
d 2V = 0 (Laplace’s eqn) dx 2 dv ⇒ =a dx ⇒ V = ax + b ------------------ (A) (where a & b are content of integration and are to be determined by boundary conditions) At x = x1 , V1 = ax1 + b And At x = x2 , V2 = ax2 + b ------------------ (i) ⇒ a( x1 − x2 ) = V1 − V2 V − V2 ⇒a= 1 ------------------ (ii) x1 − x 2
Ans:
14
AE14
ELECTROMAGNETICS AND RADIATION
V2 x1 − V1 x 2 x1 − x 2 Substituting values a & b in eqn (A) b = V1 − ax1 =
V= Q.7
------------------ (iii)
V1 ( x − x 2 ) − V2 ( x − x1 ) . x1 − x 2
State and explain Biot-Savart’s Law relating the magnetic field produced at a point due to current in a small elemental wire. (6)
Ans:Biot - Savart’s Law: Consider a differential current element of a filamentary conductor carrying a current I as shown in Fig.7.3. The law of Biot-Savart then states that at any point P the magnitude of the magnetic field intensity produced by the differential element is proportional to the product of the current, the magnitude of the differential length, and the sine of the angle lying between the filament and a line connecting the filament to the point P where the field is desired. The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differential element to the pint P. The direction of the magnetic field intensity is normal to the plane containing the differential filament and the line drawn from the filament to the point P. Of the two possible normals, that one is to be chosen which is in the direction of progress of a right handed screw turned from dL, through the smaller angle to the line from the filament to P. The Biot - Savart’s Law may be written using vector notation as I dL × a12 dH = ------------------ (7.1) 4πd 2 The unit of H is ampere I m. In terms of surface current density ( K ) and volume current density ( J ) the expression for magnetic field intensity are K dS × a12 dH = ------------------ (7.2) 4πd 2 And J dv × a12 dH = ------------------ (7.3) 4πd 2 Where point 2 is that point at which magnetic field intensity is desired, point I is the point at which differential current element is located and d is the distance between points 1 and 2. Q.8
What is a uniform plane wave? Why is the study of uniform plane waves important? Discuss the parameters ω, β and v p associated with sinusoidally time-varying uniform plane
(2 + 2 +4)
waves.
15
AE14
ELECTROMAGNETICS AND RADIATION Ans:A uniform plane wave is defined as a wave whose value remains constant throughout a plane which is transverse to the direction of propagation of the wave. For example, if the wave is travelling in a z direction, then the plane z = constant will be perpendicular to the direction of propagation. In the plane z = constant, the variables are x and y. Hence for this uniform plane wave, the electric field intensity and magnetic field intensity are independent of x and y coordinate and are the functions of z coordinates only. UNIFORM PLANE WAVE OR TRANSVERSE ELECTRO-MAGNETIC WAVE: Transformation of a time varying quantity into phasor: Let E x be the time varying quantity which is either a function of sine or cosine. Then the phasor of E x is written as E xx and is obtained by dropping Re and suppressing e jωt . If E x = E0 cos(ωt + ψ ) = E0 Re[cos(ωt + ψ ) + j sin(ωt + ψ )]
= E0 Re[e j (ωt +ψ ) ] = E0 Re[e jωt + jψ ) ] Then, E xx = E0 e jψ Where Re means the real part of the following quantity is to be taken. Further it can be dE x proved that the phasor of is given by multiplying the phasor of E x by jω , i.e. phasor dt dE of x = jωE xx . dt Transformation of phasor into a time varying quantity. To transfer a phasor into a time varying quantity, multiply that phasor by e jωt and then consider the real part (Re) of that product. If E xx = Eo e jψ , then
[
]
[
]
[
E x = Re E xx e jωt = Re Eo e jωt e jψ = Re Eo e j (ωt +ψ )
]
= Eo cos(ωt + ψ ) Maxwell’s equations in phasor form. The four Maxwell’s equations in point form or differential form are ∂B ∂H ∇×E = − = −µ. ∂t ∂t ∂D ∂E ∇×H = J + = σE + ε ∂t ∂t ∇.D = ρv
∇ .B = 0 These equations in phasor form may be written as ∇ × E s = − jωµ .H s --------------- (10.1 a) ∇ × H s = (σ + jωε ) Es
--------------- (10.1 b)
∇.Ds = ρv
--------------- (10.1 c)
∇.Bs = 0 --------------- (10.1 d) Wave equation for electric field intensity. Using the vector identity ∇ × ∇ × E s = ∇ ( ∇ .E s ) − ∇ 2 E s
16
AE14
ELECTROMAGNETICS AND RADIATION LHS = ∇ × ∇ × Es = ∇ × (− jωµH s )
[Q Using Eq 10.1a]
= − jωµ ∇ × H s = − jωµ (σ + jωε )Es
D RHS = ∇ (∇.E s ) − ∇ − 2 E s = ∇ ∇. s ε 1 = ∇ ( ρv ) − ∇ − 2 E s
[Q Using Eq. 10.1b]
− ∇ −2 E s [Q Using Eq. 10.1c]
ε
Assuming ρv is equal to zero, we get RHS = − ∇ −2 E s
∴ − ∇ −2 E s = − jωµ (σ + jωε )E s ,
or
∇ −2 E s = jωµ (σ + jωε )E s The quantity jωµ (σ + jωε ) is denoted by γ 2 . Hence
∇ −2 E s = γ 2 E s -------------------- (10.2a) This is the wave equation for electric field intensity. Similarly it can be proved that, the wave equation for magnetic field intensity is ∇ −2 H s = γ 2 H s -------------------- (10.2b) Propagation constant, Attenuation constant and Phase constant In the wave equation ∇ −2 E s = γ 2 E s The quantity γ is known as propagation constant and is defined by the equation
γ 2 = jωµ (σ + jωε ) Or γ = jωµ (σ + jωε ) The units of propagation constant is per m. It is a complex number. The real part is denoted by α and is known as attenuation constant. The unit is N p / m where N p stands for neper. The imaginary part is denoted by β
Q.9 .
Distinguish between internal inductance and external inductance. Discuss the concept of flux linkage pertinent to the determination of the internal inductance. (3 + 3)
Ans:Refer section 9.10 Page No 313 of Engineering electromagnetics by William H Hayt. Q.10
Explain the following: (i) Characteristic impedance. (ii) Distortionless line. (iii) Voltage Standing Wave Ratio (VSWR). Reflection coefficient. (iv)
(8)
Ans:Refer section 8.13, 8.14, 8.27.1, 8.28 Page No 646 of “Electromagnetic Field and Waves” by K D Prasad. Q.11
Explain the impedance transformation property of a quarter wave transmission line. 17
(7)
AE14
ELECTROMAGNETICS AND RADIATION Ans:Refer section 8.29.2, Page No 646 of “Electromagnetic Field and Waves” by K D Prasad.
Q.12
Calculate the characteristic impedance Z 0 , propagation constant γ and the line constants of an open wire loss less line of 50 Km long operating at f = 700 Hz if
Z0C = 286∠ − 40o Ω
(7)
o
ZSC = 1520∠16 Ω Ans: Z o = Z oc × Z ςc = 286∠ − 40 × 1520∠16 = 659.3∠ − 12 0 659.3 Y= . 700
Q.13.
What is dominant mode? Which one of the rectangular waveguide modes is the dominant mode? How do the dimensions of a rectangular cavity resonator determine the frequencies (4+6) of oscillation of the resonator?
Ans:Refer section 9.14 and 9.7 of “Electromagnetic Field and Waves” by K D Prasad. Q.14
A rectangular waveguide measures 3 × 4.5cm internally and has a 10 GHz signal propagated in it. Calculate the cut off frequency (λ c ) and the guide wavelength λ g .
( )
(4)
2
Ans: λ c =
2
2
m n + a b TE10 mode m = 1, n = 0 2a λc = = 2 × 0.045 = 0.090m m
λg =
λ
λ 1 − λc λg = 0.0318m .
Q.15
2
where λ =
3 ×10 8 = 0.03m 10 × 10 9
What is a Hertzian dipole? Discuss the time variations of the current and charges associated with the Hertzian dipole. Also discuss the characteristics of the electromagnetic field due to the Hertzian dipole. (2 + 3 + 3)
Ans:Refer section 10.42, Page No 850 of “Electromagnetic Field and Waves” by K D Prasad. Q.16
Deduce the radiation resistance and the directivity for a half wave dipole?
(6)
Ans:Refer section 10.47, Page No 855 of “Electromagnetic Field and Waves” by K D Prasad. 18
AE14 Q.17
ELECTROMAGNETICS AND RADIATION Distinguish between broadside and end fire radiation patterns with suitable sketches. What is an array factor? Provide a physical explanation for the array factor. (4 + 4)
Ans:Refer section 10.3, Page No 781 of “Electromagnetic Field and Waves” by K D Prasad. Q.18
Estimate the maximum usable frequency (MUF) for a critical frequency of 10 MHz and an angle of incidence of 45o .
(6)
Ans: MUF = fc sec θ = 10MHz × sec 45 0 = 10 2 MHz
Q.19
What are the boundary conditions for static electric fields in the general form at the interface between two different dielectric media? Explain. (8)
Ans:At the boundary between two dielectric media, the tangential and the normal component electric displacement have to be analysed. (i) Tangential component: Let E1 & E2 are the permittivities for two dielectric media. Et1 & Et2 are the tangential components in media I & media II. A rectangular closed path abcd has been considered. For the closed path
∫
b
c
d
a
a
b
c
d
E.dl = ∫ E.dl + ∫ E.dl + ∫ E.dl + ∫ E.dl = 0
Let ab = cd ⇒ 0, at the boundary So φE.dl = 0 + Et1 cd + 0 − Et1 cd = 0 So Et1 = Et2
(ii)
Normal components of electric displacement A small cylindrical pill box of height ∆h and crosssectional area ∆s has been considered. We know that ∫ D.ds = Q enclosed s
Or
∫ D.ds + ∫ D.ds + ∫ D.ds = Q top
bottom
sides
19
(enclosed)
AE14
ELECTROMAGNETICS AND RADIATION Let ∆h → 0 at the boundary and ρ s be the surface charge density enclosed at the surface of boundary. Also, Dn1 & Dn2 are the normal components of D. So, Dn1 ∆s − Dn2 ∆s + 0 = ρ s ∆s
Q Q = ρ s ∆s
Or Dn1 − Dn2 = ρ s For free surface charge density, ρ s = 0 Dn1 = Dn2 , for ρ s = 0 .
Q.20
Using Gauss’s law in integral form, obtain the electric field due to following charge distribution in spherical coordinates: K ρ(r, θ, φ ) = r 2 for 0 < r < R Charge density, for R < r < ∞ 0 Where, R is the radius of spherical volume, K is constant & r is radial distance. (8)
Ans:For r < R, dQr = ρdv K = 2 4πr 2 dr r = 4πKdr So, total charge enclosed within radius r (r < R) Q r = ∫ dQ r = ∫ 4πKdr = 4πKr ------------------------- (1) 20
AE14
ELECTROMAGNETICS AND RADIATION From Gauss’s law, (r < R) ∈0 ∫ E.ds = ∫ ρdv ∈0 E 4π/r 2 = 4π/Kr K Or, E = i r , for r < R ∈0 r At r = R, R
∈0
∫ E.ds = ∫ ρdv 0 R
=
K
∫r
2
4πr 2 dv
0
Or, ∈0 E 4π/r 2 = 4π/ KR K E= i r , for r = R ∈0 R For r > R r R R K ∈0 ∫ Eds = ∫ ρdv = ∫ 2 4πr 2 dr 0 0 0 r Or, ∈0 E 4π/r 2 = 4π/ KR K R E= ir for r > R. ∈0 r 2
Q.21
State Poisson’s equation. How is it derived? Using Laplace’s equation, for a parallel plate capacitor with the plate surfaces normal to X-axis, find the potential at any point between (2+4+4) the plates. Given, V = V1 at x = x1 and V = V2 at x = x 2 .
Ans:From Maxwell’s first equation i.e. Gauss’s Law in differential form, we have r r r r ∇.D = ρ D =∈ E r r Using E = −∇V Where ∈ is the permittivity of the medium. We obtain − ∇. ∈ ∇V = ρ Using vector identity ∇.φA = φ∇. A + A.∇φ We get r r r r ∈ ∇.∇V + ∇ ∈ .∇V = − ρ Or ∈ ∇ 2V + ∇ ∈ .∇V = − ρ Or ∇ 2V = −
ρ
as ∇ ∈= 0 for uniform ∈. ∈ This is Poisson’s equation which governs the relationship between the volume charge density ρ in a region of uniform permittivity ∈ to the electric scalar potential V in that region. In carlesian coordinate. Poisson’s equation becomes
21
AE14
ELECTROMAGNETICS AND RADIATION
∂ 2V ∂ 2V ∂ 2V ρ + 2 + 2 =− 2 ∂x ∂y ∂z ∈ If V is a function of x only then we have ∂ 2V ρ =− 2 ∂x ∈
Laplace equation for a parallel plate capacitor is ∂ 2V =0 ∂x 2 d 2V =0 dx 2 On integration, ∂V =a ∂x And again after integration V = ax + b ------------------ (1) Where a & b are constants of integration to be determined by boundary condition. Boundary conditions are V = V1 at x = x1 V = V2 at x = x2 We then have V1 = ax1 + b ------------------ (2) V2 = ax2 + b V1 − V2 V x −V x and b = 2 1 1 2 x1 − x2 x1 − x2 Putting the value of a and b in equation (1), we obtain V ( x − x2 ) − V2 ( x − x1 ) V= 1 x1 − x2 Which gives potential at any point x between the plates i.e. x1 < x < x2 .
Equation (2) ⇒ a =
Q.22
State and explain Biot-Savart’s law.
(2+4)
Ans:The Biot-Savart Law states that the magnetic flux density due to a current element Idl(current I passing from an infinite-simal small length of conductor, dl) in free space is given by 22
AE14
ELECTROMAGNETICS AND RADIATION
r µ I dl × iˆ R dB = 0 4π R 2
Where µ 0 is the permeability of free space which is equal to 4π × 10 −7 . R = distance from the element to the point P in meters. iˆR = a unit vector directed from element to P. The above equation can also be expressed as µ Idl dB = 0 2 sin α .iˆφ 4πR Where α is the angle between the current element and the line joining it to p measured in the same order dl to iˆR . The direction of dB is perpendicular to the plane ( dl × iR ) containing the element and the line joining the element to P as given by cross product ( dl × iˆR ).
(
)
For α = 0, i.e. along a line in alignment with the element, no field ( dB = 0 ) is produced and for α = 90 0 i.e. along a line perpendicular to the element. Also the field is inversely proportional to R 2 .
Q.23
What is Poynting vector? How is the Poynting theorem derived from Maxwell’s curl equations? Explain Poynting theorem. (2+8) r r Ans: The quantity E × H as the power flow density vector associated with the electromagnetic field. It is known as the poynting vector after r the name of J. H. Poynting and is denoted by the symbol P . Thus, r r r P = E×H r r It is instantaneous poynting vector, since E and H are instantaneous field vectors. Poynting theorem: From vector identity r r r r r r rr r ∇. E × H = H .∇ × E − E.∇ × H ------------------------ (1) And Maxwell’s curl equations, r r r r ∂B ∂H ∇× E = − = −µ ∂t ∂t ------------------------ (2) r r r r r ∂D r ∂E ∇ × H = Jc + = σE + ∈ ∂t ∂t Using Maxwell’s curl equations (2) in equation (1), we have
(
)
23
AE14
ELECTROMAGNETICS AND RADIATION
r r r r r r ∂H r r ∂E − E. σE + ∈ ∇. E × H = H . − µ ∂ t ∂ t r r r ∂H r r r ∂E = − µH . − σE.E − ∈ E. ∂t ∂t r r r r ∂ 1 ∂ 1 r r = − µ H .H − σE.E − ∈ E.E ∂t 2 ∂t 2 ∂ 1 ∂ 1 = − σE 2 − ∈ E 2 − µH 2 ----------------------- (3) ∂t 2 ∂t 2 r r r Substituting P for E × H and taking the volume integral of both sides of equation (3) over the volume V, we obtain ∂ 1 ∂ 1 2 2 2 ∫v (∇.P )dv = −∫v σE dv − ∫v ∂t 2 ∈ E dv − ∫v ∂t 2 µH dv Using divergence theorem on left hand side we have r r ∂ 1 ∂ 1 2 2 2 P -------------------- (4) ∫s .ds = −∫v σE dv − ∂t ∫v 2 ∈ E dv − ∂t ∫v 2 µH dv Where s is the surface bounding the volume v. The equation (4) is called the pointing theorem in which The power dissipation density, Pd = σE 2 1 The electric stored energy density, we = ∈ E 2 2 1 The magnetic stored energy density, wm = µH 2 2 And, the power flowing out of the closed surface s i.e. LHS of eqn (4) = ∫ P.ds .
(
)
s
Q.24
The conduction current density in a lossy dielectric is given as J C = 0.02 Sin 109 t A m 2 , 3 find the displacement current density if σ = 10 mho
m
, ∈r = 6.5 & ∈o = 8.854 × 10 −12 .
Ans: J c = σE = 0.02 sin 10 9 t
(6)
= 103 E So, E = 2 × 10 −5 sin 109 t The displacement current density is given by ∂D ∂ Jd = = ∈0∈r 2 × 10 −5 sin 109 t ∂t ∂t = 8.854 × 10 −12 × 6.5 × 2 × 10 −5 × 109 cos 109 t Thus, J d = 1.15 × 10 −6 cos 109 t A/m.
(
)
→
Q.25
Derive general expressions for reflection coefficient and transmission coefficient for E and →
H fields when an electromagnetic wave is incident normally on the boundary separating two different perfectly dielectric media. (8)
24
AE14
ELECTROMAGNETICS AND RADIATION Ans:When a plane electromagnetic wave is incident normally on the surface of a perfect dielectric, part of energy is transmitted and part of it is reflected A perfect dielectric is one with two conductivity so there is no loss of power in propagation through the dielectric. In the figure suffix ‘i’, ‘r’ & ‘t’ signify the “incident”, “reflected” and “transmitted” components respectively for E and H. Intrinsic impedances For medium 1, η1 =
µ1 ∈1
& for medium 2, η 2 =
µ2
∈2 The relationships for electric and magnetic fields are Ei = η1 H i ------------------ (1) Er = −η1 H r ------------------ (2) (energy direction changed, so –ve sign) And Et = η 2 H t ------------------ (3) The continuity of components of E & H require that Ei + Er = Et ------------------ (4) Hi + H r = Ht (1) & (2) ⇒ H i + H r =
1
η1
( Ei − E r ) = H t =
1
η2
(Ei + Er )
Or η 2 (Ei − Er ) = η1 (Ei + Er )
------------------ (5) E η −η Using eqn (4) & (5), the reflection coefficient, R = r = 2 1 Ei η 2 + η1 E E + Er And transmission coefficient, T = t = i Ei Ei E 2η 2 = 1+ r = Ei η1 + η 2 2η 2 Or T = η1 + η 2 Similarly, we can get reflection and transmission coefficients for H as H η −η R= r = 1 2 H i η1 + η 2 2η1 H ηE And T = t = 1 t = H i η 2 Ei η1 + η 2
Q.26
Consider the volume current density distribution in cylindrical coordinates as J (r, φ, z ) = 0, 0 < r < a r = J o ˆi z , a < r < b a = 0, b < r < ∞ 25
AE14
ELECTROMAGNETICS AND RADIATION Where a and b are inner and outer radii of the cylinder. Find the magnetic field intensity in various regions. (8)
Ans:For region: 0 < r < a, J(r, ¢, z) = 0 Using Ampere’s law, a 2π r r ∫ H .dl = ∫ J .ds = ∫ ∫ J rdrdφi z = 0 r =0 φ = 0
Or H.2πr = 0 Or H = 0 for region I (0 < r < a) r Region II, a < r < b, J = J 0 iˆz a
Applying Ampere’s Law,
∫
2π
r
r
J r3 r H .dl = ∫ ∫ J 0 rdrdφ = 0 φ a a 3 φ =0 r = a a
Or, H .2πr =
2π J 0 r 3 − a3 3a
(
2π 0
)
J0 3 r − a 3 iˆφ 3ar J At r = b, H = 0 b 3 − a 3 iˆφ 3ab Region III, b < r < ∞ , J = 0 2π r = b J r We have, ∫ H .dl = ∫ ∫ 0 rdrdφ φ =0 r = a a Or, H =
(
)
(
Or, H =
Q.27
)
J0 3 b − a 3 iˆφ . 3ar
(
)
Calculate the electric field due to a line charge considering it a special Gaussian surface.
(5) Ans:Assuming line charge along z-axis Electric field E or flux density D can only have an r component can only depend on r. A closed right circular cylinder or radius r, whose axis is z-axis, has been assumed for the purpose of Gussian surface. From Gauss’s Law, 26
AE14
ELECTROMAGNETICS AND RADIATION Q=
∫ D.ds + ∫ D.ds + ∫ D.ds s =1
s=2
s =3
= 0 + 0 + ∫ ∈ E.ds s =3
[as on the surface 1 & 2, E & ds are far so E.ds = 0.] ------------------ (1)
= ∈ E 2πrL
If ρ is the charge per unit length and the length of Gaussian Cylindrical surface is, L, Q ρ= L (1) ⇒ ρL =∈ E 2πrL ⇒ E =
Q.28
ρ ir 2πr ∈
Find out the magnetic field intensity at any point due to a current carrying conductor of finite length using the Biot-Savart law. (5)
Ans:dH =
Idl × i R 4πR
2
=
Idl sin(90 + α ) iφ 4πR 2
27
AE14
ELECTROMAGNETICS AND RADIATION
Now Rdα = dl cos α Rdα dl = cos α dα IR . cos α α cos So dH = .iφ 4πR 2 Idα Idα = .iφ = 4πR r 4π cos α α 1 2 Or, H = cos α dα .iφ 4πR α∫1 Or, H =
Q.29
1 (sin α 2 − sin α1 ).iφ . 4πR
Find the characteristic impedance, propagation constant and velocity of propagation for a transmission line having the following parameters: R = 84 ohm/Km, G = 10 −6 mho/Km, L=0.01 Henry / Km, C=0.061 µF /Km and frequency = 1000 Hz.
Ans:Characteristic impedance, Z R + jω L Z0 = = Y G + jωC Where Z = R + jωL = 84 + j 2π 1000 × 0.01 = 84 + j 62.83 = 104.9∠36.80 ohm/km 28
(6)
AE14
ELECTROMAGNETICS AND RADIATION And Y = G + jωC = 10 −6 + j 2π × 1000 × 0.061× 10 −6 = 10 −6 + j 383.27 × 10 −6 = 383.27 × 10 −6 ∠89.850 mho/km Z 104.9∠36.80 ∴ Z0 = = Y 383.27 × 10 −6 ∠89.850 = 523.16∠ − 26.530 ohm. Propagation Constant γ = ZY = ( R + jωL)(G + jωC )
= 104.9∠36.80 × 383.27 × 10 −6 ∠89.85 = 0.201∠63.330 = 0.09 + j 0.18 per km = α + jβ . So β = 0.18 rad/km ∴ Velocity of propagation, ω 2π × 1000 vp = = = 34.9 × 10 −6 m/sec. −3 β 0.18 × 10
Q.30
How does dissipation-less transmission lines act as tuned circuit elements? Explain.
(9)
Ans:At high frequencies (300 MHz to 3 GHz ), a transmission line can be used as circuit element (like a capacitor or an inductor). At these frequencies the physical length of the line is convenient to use. For lossless line, γ = jβ and Z 0 = R0 And tan hγl = j tan β l . The input impedance of the line Z in with load Z L becomes
Z + JR0 tan βl Z in = R0 L ---------------------- (1) R0 + jZ L tan βl (i) Open circuited line (Z L → ∞ ) R0 Equation (1) ⇒ Z in = − j = − jR0 cot β l ---------------- (2) tan β l = jX 0 (purely reactive) So, the input impedance of an open circuited lossless line is purely reactive, i.e. the line can be either capacitive or indirective depending on the value of β l . Now ( X 0 = − R0 cot β l ) vs l is plotted as given below: X 0 can vary from − ∞ to + ∞ . For a very short length of the line βl λ 0 .
(iii)
For TE01 mode i.e. m =0, n = 1, and λc = 2b = 2 × 3.44 = 6.88 cm. Hence this mode will not propagate because λ c < λ 0 .
Obviously the higher TE mode will not propagate for λc < λ0 for other values of m & n. Also for TM mn mode the lowest value of m & n is unity hence no TM mode is possible at the frequency. λ0 0 .1 Since the guide wave length λg = = = 0.141 metres. 2 2 0 . 1 λ 1− 1 − c 0.144 λ0
λg 0.141 We get phase velocity v p = c = × 3 × 108 = 4.23 × 108 m/sec. 0.100 λ0 λ 0.100 And group velocity, vg = 0 c = × 3 × 108 = 2.28 × 108 m/sec. λ 0 . 141 g 2π 2 × 3.14 The phase constant, β = = = 44.7 . λg 0.141 Q.36
Using Gauss theorem, derive an expression for the electric field intensity due to a line charge of infinite length at distance R. (8)
Ans:
R
Figure dΨ =
+ + + + + + + + + + +
D. ds ^
= =
nˆ
^
D ds r .n D ds
(rˆ & nˆ are parallel ) 35
AE14
ELECTROMAGNETICS AND RADIATION ^
^
where r is unit vector in the direction of flux density D and n in the direction of normal drawn to the surface. Total flux from lateral surface. ψ L = ∫ dψ L s
=
^ ^ Dds cos 0 r & n are parallel ∫s
= D ∫ ds s
= D (2 πRl) Total flux from Top surface ψ T = ∫ dΨT s ^ ^ = ∫ Dds cos 0 r & n are parallel s ψT = 0 Total flux from bottom surface Ψ B = ∫ dΨ B s
=
^ ^ 0 Dds cos θ θ = 90 , because r & n are ⊥ er to each other ∫s
ψB = O Therefore total flux coming one of the cylinder is given by. ψ = ψL + ψT + ψB ψ = D 2 π Rl By Gauss law ψ = Q, total charge enclosed Q = D 2 π Rl Q ρ L l = D 2 π Rl ρ= l ρ L l = εΕ2π R l D = ε E
(
E=
)
ρL ^ r 2πεR ^
where r is a unit vector normal to the line charge.
Q.37
A positive charge density of Qv C/m3 occupies a solid sphere. At a point in the interior at a distance ‘r’ from the center a small probe charge of +q is inserted. What is the force acting on the probe charge? (8)
Ans: Q 1 = Q∨ X 4 π r 3 3 Total flux over the surface = charge enclosed. 1 ∫ D . ds = Q s
36
AE14
ELECTROMAGNETICS AND RADIATION ^ ^ 0 = Dds cos θ θ 0 , because r & n are parallel ∫s
∫ Dds cos 0 = Q
1
s
D 4 π r2 = Q1 D 4 π r 2 = Qν X 4 πr 3 3 2 εΕ 4π r = Qν X 4 3 π r 3 Q r Ε= V 3ε Since E =
F q
Q r Force on the probe charge = q V newton 3ε
Q.38
Prove Ampere’s circuital law for time varying field condition in differential form.
(7)
Ans: ∇. D = ρ ∂ρ ∂ ∇D = V ∂t ∂t
( )
∂ ρ ∂ρ V = ∇. ∂t ∂ t ∂ρ ∇ . J = − V {equation of continuity} ∂t ∂D ∇ . J = − ∇. ∂t ∂ D ∇. J + = 0 ∂ t ∂ D For time varying case ∇ J = 0, is changed to ∇. J + = 0. So ∂ t
∂D . ∂t ∴ Ampere' s circuital law in po int form becomes J+
∇X H
Q.39
=
J+
∂D ∂t
A circular conductor of 1cm radius has an internal magnetic field
37
J must be replaced by
AE14
ELECTROMAGNETICS AND RADIATION
1 1 r 2 sin ar − cos ar i φ r a a where a = π/ 2 ro and ro is the radius of the conductor. Calculate the total current in the conductor ( i φ is unit vector). (9)
H=
dΦ
dl ro
∫ H . dl = I
Ans:
r = ro dl = ro d φ Hφ =
1 1 ro a 2
∫ H.
dl = ∫ H φ ro d φ = I
sin ar0 −
r cos aro a
1 1 ro sin aro − cos aro o a2 a
∫r
π
a =
∴
∫
2 ro 1
∫ a
4ro
2
π2
4ro2
π 4 ro 2
π2 I = r =
2
}
} ro dφ = I
by data π ro π sin ro − cos ro dφ = I 2 ro 2ro a
( 2)= 0
dφ = I because π
∫ dφ = I (2π ) = I 8ro2
π 10 −2 m
38
AE14
ELECTROMAGNETICS AND RADIATION I = 2.55 × 10 −4 A
Q.40
Prove the electric field normal components are discontinuous across the boundary of separation between two dielectrics. (6)
∫ D1 . ds
Ans: Flux over top surface = Flux over bottom surface =
∫D
2
.ds
Since height of the box is vanishingly Small the flux through the sides of box is ignored Net flux =
∫D
1
∫D
ds −
2
. ds
r Net flux = D1 ∆s − D2 ∆s Q From gauss law net flux = total charge which is equal to ρ s ∆s because ρ s = ∆s
ρ s ∆s = ( D .nˆ1 − D2 .nˆ 2 )∆s ρ s ∆s = (Dn1 − Dn 2 )∆s Dn1 − Dn 2 = ρ s ε1En1 − ε 2 En 2 = ρ s ) n
→ D1
ε1 ε2
→ D2
) n2
Since it is dielectric-dielectric interface ρ s = 0
ε 1Ε n1 − ε 2 Ε n 2 = 0 Since E n1 ≠ E 2 Q.41
Measurement made in the atmosphere show that there is an electric field which varies widely from time to time particularly during thunderstorms. Its average values on the surface of the earth and at a height of 1500m are found to be 100 V and 25 V m m
39
AE14
ELECTROMAGNETICS AND RADIATION directed downward respectively. Using Poisson’s equation calculate i) the mean space charge in the atmosphere between 0 and 1500 m altitude ii) surface charge density on earth. (10)
Ans: ∇ 2 V = ρ V= −ρ
εo
ε 0 ∫ xdx + ∫ Adx
x2 ε 0 2 + Ax + B Using V = 0 at X = 0
V = −ρ
We get constant B = 0
Ex = p
ε o x + 100
At x = 1500m, E = 25 v 25 =
ρ
m
ε o x 1500 + 100
25 − 100 x(8.854 x10 −12 ) 1500 ρ = − 4.427 x 10 −13 c 3 m To find surface charge density on cases
ρ =
E
ρs 2ε o
ρ s 2 X 8.854 X 10−12 X 100 = 1.771 x 10-9 c
Q.42
Derive an significance.
m2
expression
for
equation
Ans: ds
Let the charge in elementary volume = ρ v Entire charge within the closed surface = ∫ ρ v dv v
40
of
continuity
and
explain
its (6)
AE14
ELECTROMAGNETICS AND RADIATION When there is a charge flow out of the surface, then, with time there will be reduction of charge −∂ within the surface given by ρ v dv ∂t ∫v
∫ s
∫
r r −∂ J . ds = − ρ v dv ∂t ∫v r r r ∇.Jdv = φsJ .ds
v
r
−∂ ρ v dv ∫ ∂ t v v r ∂ρv .∇.J = − ∂t r Significance : Divergence of current density J represents the net charges which flows out of the enclosed surface and it will be equal to reduction of charge within the surface with respect to time.
∫ V .J dv =
Q.43
Do the fields
E = E m sin x sin t a y and H =
equation?
Ans: ∇ x E = xˆ
Em cos x cos t a z , satisfy Maxwell’s m (10)
− ∂B ∂H = − µ0 ∂t ∂t
yˆ zˆ
∂ ∂ ∂ ∂ = - µo H x xˆ + H y yˆ + H z zˆ ∂x ∂y oz ∂t Ex Ey Ez
[
]
Since wave is traveling in x direction, Ex, Hx components are absent as well ∂ produce null result. Hence ∂z ∂Ey ∂Hy ∂Ez ∂Hz zˆ = − µo yˆ − µo yˆ + zˆ ∂x ∂t ∂x ∂t Equating the components along z-direction ∂Ey ∂Hz = − µo → (1) ∂x ∂t r Since Ey= E = Em sin x sin t aˆy ( from data ) ∂Ey = Em cos x sin t → ( A) ∂x r Em Hy = H = Cosx cos t aˆz ( from data )
µo
41
∂ and ∂y
AE14
ELECTROMAGNETICS AND RADIATION If wave is travelling in x-direction Hx and Ex are absent, as well
∂ ∂ and produces null ∂y ∂y
result, hence ∂Hz ∂Hy ∂Ey ∂Ez - yˆ + zˆ = ε yˆ + ε zˆ ∂x ∂x ∂t ∂t Equating components along y- direction ∂Hz ∂Ey =ε → ( 4) ∂x ∂t Similarly for z-direction ∂Hy ∂Ez =ε ∂x ∂t Solution for the wave equation for a travelling in x-direction is given by Ey= f(x-vt) 1 V=
µε
Let u = (x-vt) Ey=f1(µ) ∂Hz − E m = cos x sin t µo ∂t −µ
∂Hz = Em cos x sin t → ( B ) ∂t
From equation (A) & (B) ∂Ey ∂Hz = − µo ∂x ∂t Thus the fields satisfy the maxwell’s equations.
Q.44
Write short notes on skin depth and skin effect r r r ∂E Ans: ∇xH = J + ε ∂t It wave is travelling in free space where there are no free charges r r ∂E ∇ XH = ε ∂t xˆ yˆ zˆ
∂ ∂ ∂ ∂ r ∂x ∂y ∂y = ε ( E X xˆ + EY yˆ + E Z zˆ ) ∂t H X HY H Z
42
(4)
AE14
ELECTROMAGNETICS AND RADIATION ∂Ey ∂f 1 (u ) ∂f1 (u ) ∂u = = ∂t ∂t ∂u ∂t ∂Ey = −Vf11 ∂t Substituting this in equation (A) - ∂Hz = − ε
1 µ f1 ∂x
ε f 11 ∂x + C ∫ µ
Hz =
Differentiating Ey = f1(u) with respect to x we get ∂Ey ∂f 1 (u ) ∂f 1 (u ) ∂u ∂u = = = f 11 ∂x ∂x ∂u ∂x ∂x ∂u =1 ∂x ∂Ey ∴ f11 = ∂x hence Hz =
ε
µ Ey+C
or Ey = µ ε Hz → ( B ) Similarly we can prove
µ H → (C ) ε Y
EZ = -
Adding B & C after squaring we get Ey2+Ez2 = µ E = H
µ
2 2 ε (H Y + H Z )
ε
r because E = Ey 2 + Ez 2 r & H = Hy 2 + Hz 2
Q.45
Prove for a travelling uniform plane wave µ E = ε H →
where E and H are amplitudes of
→
E and H , respectively.
43
(12)
AE14
ELECTROMAGNETICS AND RADIATION Ans:
→ E
δ
x
When an electromagnetic wave enters a conducting medium its amplitude decreases exponentially and becomes practically zero after penetrating a small distance. As a result, the current induced by the wave exists only near the surface of the conductor. This effect is called skin effect The skin depth is defined as the depth of a conductor at which the amplitude of an incident wave drops (1/e) times its value of amplitude at the time of incidence. The depth of penetration expression is given by 2 1 = δ= ωµσ πfµσ
Q.46
Discuss the following terms (i) Wavelength (iii) Group velocity
(ii) Phase velocity (iv) Propagation constant
(8)
Ans:Wave length : It is defined as the distance in which the phase change of 2π radians is effected by a wave travelling along the line. 2π λ=
β
Phase velocity : It is defined as the velocity with which a signal of single frequency propagates along the line at a particularly frequency f. Vp= ω km β sec Group velocity : If the transmission medium is such that different frequencies travel with different velocities, then the line or the medium is said to be dispersive. In that cases, signals are propagated with a velocity known as group velocity , Vg ω − ω1 Vg = 2 β 2 − β1 Propagation constant : It reveals the nature in which the waves are propagated along the line that is the manners in which the voltage v and current I vary with distance x. I P = log e s IR OR 44
AE14
ELECTROMAGNETICS AND RADIATION
P = log e Q.47
Vs VR
A high frequency transmission line consists of a pair of open wires having a distributed capacitance of 0.01 µF per Km and a distributed inductance of 3mH per Km. What is the characteristic impedance and propagation constant at f=10MHz? (4) L = C IPI = ω LC
Ans:Zo =
3 X 10 3 = 547.7 ohms 0.01X 1 0 6
= 2πX 10 X 10 6 (0.01) X (1 0 6 ) X (3) X (1 0 3 ) = 344.156 rad km
Q.48
(4)
Comment on Impedance matching device.
Ans:Impedance matching device : According to maximum power transfer theorem, for maximum power transfer source or generator impedance real part to be equal real part of load impedance and generator impedance imaginary part to be conjugate of imaginary part of load impedance. When load impedance are fixed, one has to employ matching network between source and load to match the impedance of source and load. These networks are called matching impedance devices. For example quarter wave transformer is a impedance matching device. Q.49
Derive the condition for a distortionless line and comment on the result.
(8)
Ans:
{
)}
1 RG − ω 2 LC + RC − ω 2 LC ) 2 + ω 2 ( RC + LG ) 2 2 In order to eliminate frequency distortion the attenuation constant must be made independent of frequency.
α=
(
Thus the desired condition ( RC − ω 2 LC ) 2 + ω 2 ( RC + LG ) 2 = w2LC+K Squaring on both and solving, we get R2G2+ω2(R2C2+LG2) = K2+2ω2LCK Which implies K = RG And 2LCK = R2C2+LG2 R2C2-2LCRG+L2G2= 0 (RC – LG)2 = 0 L C = R G
Substituting this in the ‘α’ equation we get α = 45
RG
AE14
ELECTROMAGNETICS AND RADIATION
[(
]
1 ω 2 LC − RG + RG − ω 2 LC ) 2 + ω 2 ( RC + LG ) 2 2 If phase Vp to be independent of frequency β must become function of frequency
β=
)
ω 2 (2 K 2 − LC ) + RG =
(RG − ω
2
LC
)
2
+ ω 2 ( RC + LG )
Squaring and rearranging we get (2K 2 -LC)2 = L2C2 K= LC And 2 RG (2K 2 LC)= R2C2+L2G2 (RC – LG)2 =0 RC = LG
Comment : Perfect transmission line is not practically possible. A well constructed transmission line has very small value of G, which requires very large α in order to satisfy RC = LG. If G is increased attenuation is increased . Resistance R cannot be reduced, because the diameter of conductor increases and it becomes costlier affair. Thus increasing the value of L is the only solution. Q.50
An antenna array presents an impedance of 300 ohms to the transmission line feeding it. The transmission line consists of two open wire lines whose spacing is 9” apart and the diameter of the wire is 0.1”, calculate the dimension of a quarter wave line required for (8) matching.
Ans:The characteristic impedance of the feeder is given by Zo = 276 log10 s
r
S = Spacing between two wires R = Radius of the wire Zo = 276 log10 9 0.05 = 704.6 ohms Z0 = Z s Z R Z0= 704.6 X 300 Z0= 459.2 ohm Practically spacing of quarter wave matching line would be same as that of the main line. Hence, if r be the radius of the line, then Z0 = 276 log10 9 r 9 Z 459.2 Log = 0 = r 276 276 r = 0.1950 Diameter = 2xr = 0.3900 = 0.39 “ answer
Q.51
Derive the wave equation for a TE wave and obtain all the field components in a rectangular wave guide. (11)
46
AE14
ELECTROMAGNETICS AND RADIATION Ans:For a TE wave Ez = 0, Hz ≠ 0 ∂H z ∂ 2 H z + + h2H z = 0 2 2 ∂x ∂y H z = XY where x is a pure function of x only and Y is a pure function of y only Hz = (C1 cos B x + C2 sin Bx) (C3 cos Ay + Cy sin Ay) Where C1,C2,C3, C4 are constants and are evaluated using boundary conditions. Boundary conditions are 1. Ex = 0 at y = 0 ∀x → 0 to a 2. Ex= 0 at y = b ∀x → 0 to a 3. Ey= 0 at y = 0 ∀x → 0 to b 4. Ey = 0 at x = a ∀y → 0 to b We know γ ∂E z jωµ ∂H z Ex = - 2 h ∂x h 2 ∂y − jωµ {(C1 cos B X + C 2 Sin B X )(− AC 3 Sin Ay + AC 4 cos Ay )} Ex = h2 Substituting the first boundary condition C4 = 0, hence Hz = (C1 cos B x + C2 sin B x ) (C3 cos Ay) γ ∂E z jωµ ∂H z We also know Ey = 2 + 2 ∂x h ∂y h jωµ Ey = 2 [− BC , sin Bx + BC 2 cos Bx][c 3 cos Ay ] h Substituting the second boundary condition C2=0 hence, Hz=C1 C3 cos B x cos Ay Again − γ ∂Ez jωµ ∂Hz Ey= 2 + 2 ∂x h ∂y h Differentiating Hz and then substituting we get jωµ Ey = − 2 C1 C 3 B sin B x cos Ay h Substituting 3rd boundary condition mπ B= a − γ ∂E z jωµ ∂H z Again, Ex = 2 − 2 ∂y h ∂x h
47
AE14
ELECTROMAGNETICS AND RADIATION jωµ C1 C 3 A cos B x sin Ay h2 Substituting 4th boundary condition nπ A = b mπ nπ ∴ Hz = C1 C3 cos X cos y a b mπ nπ jωt − r z Hz = C" cos X cos y e a b Knowing Hz, the field components of TE wave are. − γ ∂E z jωµ ∂H z Ex = 2 − 2 ∂y h ∂x h jωµ nπ mπ nπ jwt − rz Ex = C" cos X sin ye 2 h b a b − γ ∂E z jωµ ∂H z Ey = 2 + 2 ∂x h ∂y h − jωµ mπ mπ nπ jωt −γz Ey = C" sin X cos ye 2 h a a b
Ex =
Hx
mπ mπ nπ jwt − rz C" sin X cos ye h b a b
γ
2
And Hy =
Q.52
mπ mπ nπ jwt − rz C" sin X Sin ye h b a b
γ
2
Find the cut-off wavelength in a standard rectangular wave guide for the TE11mode.
(5) Ans: b= a
2
2
λc =
2
=
2
=
2 m + nb a 2a
( )
2
n m + a a 2
2
m 2 + 4n 2 For TE11 mode m = 1, n = 1 λc = 0.8944 a meters Q.53
r A long cylinder carries a charge of density ρ = kr , find E inside the cylinder.
48
(6)
AE14
ELECTROMAGNETICS AND RADIATION Ans: E 2πrl =
2 πkl 3 r 3 ∈0
r 1 kr 2 rˆ Or E = 3 ∈0 r q in E ∫ .ds = ∈0 r
Where qin = ∫ ρ dv = ∫ kr.rdr dφ dz 0
=
Q.54
2 πklr 3 3
(9)
Enunciate and give a simple proof for Ampere’s circuital law.
Ans: Refer Text Book – II, Chapter – 6. Q.55
A closed current loop having two circular arcs (of radii ‘a’ and ‘b’) and joined by two radial lines as shown in Fig.1. Find the magnetic field ‘B’ at the centre O.
(7) Ans:The magnetic field at ‘O’ due to inner arc AB is r θ µ 0i B1 = (upward) 2π 2a And the magnetic field at ‘O’ due to circular arc DC is r θ µ 0i B2 = (inward) 2π 2b Thus the resultant magnetic field at ‘O’ is µ iθ (b − a ) B = B1 − B2 = 0 4πab Q.56
What are the speed, direction of propagation and polarisation of an electro-magnetic wave whose electric field components are given as E x = 4E o cos(3x + 4 y − 500 t )
E y = 3E o cos(3x + 4 y − 500t + π) Ez = 0
(6+2+4) 49
AE14
ELECTROMAGNETICS AND RADIATION Ans: ∇ 2 E x −
1 ∂ 2 Ex =0 2 2 vx ∂t
------------------------ (1)
2 1 ∂ Ey And ∇ E y − 2 =0 2 v y ∂t 2
------------------------ (2)
Now E x = 4 E0 cos(3 x + 4 y − 500t ) ∴ ∇ 2 E x = −36 E x ∂ 2 Ex = −4(500) 2 E x 2 ∂t Substituting there in equation (1) 4 − 36 E x + 2 (500) 2 E x = 0 vx 500 vx = 3 Also E y = 3E0 cos(3 x + 4 y − 500t + π ) And
So ∇ 2 E y = −48E y And
∂2Ey ∂t
2
(2) ⇒ v y =
= −3× (500) 2 E y 500 4 2
∴ velocity, v = v x − v y
2
= 2.08 × 10 −2 m/s. The divertion of propagation is perpendicular to z axis and in x – y plane. Ey 3 =− Polarization, Ex 4 3 ∴ φ = tan −1 4
Q.57
A plane electromagnetic wave propagating in the x-direction has a wavelength of 5.0 mm. The electric field is in the y-direction and its maximum magnitude is 30 V m . Write suitable equations for the electric and magnetic fields as a function of x and t. (4) x x Ans: E = E0 sin w t − ; B = B0 sin w t − c c 2πc We have w = 2πν =
λ
2π (ct − x) = (30v / m) sin (ct − x) λ 5mm E 30v / m MaxB, B0 = 0 = = 10 −7 T C 3 × 108 m / s E = E0 sin
2π
50
AE14
ELECTROMAGNETICS AND RADIATION 2π So B = B0 sin (ct − x) λ 2π = 10 −7 T sin (ct − x) . 5.0mm
(
Q.58
)
r Derive a general expression for reflection coefficient and transmission coefficient for E and r H fields when an electromagnetic wave is incident parallel (oblique incidence) on the boundary separating two different perfectly dielectric media and also find the expression for normal incidence. (8+4)
′ = H 02 Ans: H 01 + H 01 (E01 − E01′ ) cos θ i = E02 cos θ t
But H 01 =
∈1
µ1
E01 +
∈1
µ1 ∈1
µ1
∈2 E01 ′ = E01 IIl
µ2 ∈2
µ2 E02 = E01 IIl
∈2
′ = E01
cos θ i − cos θ i + 2
∈2
µ2
∈1
µ1
E02 ∈1
µ1 ∈1
µ1
cos θ t cos θ t
cos θ i ∈1
cos θ t µ1 For normal incidence θ i = θ t = 0 Er η 2 − η1 Et 2η1 = and = . Ei η 2 + η1 Ei η 2 + η1
µ2
cos θ i +
51
AE14 Q.59
ELECTROMAGNETICS AND RADIATION Find the Laplace equation in spherical polar coordinates for V = Va at r = a =0
(4)
r=∞
1 ∂ 2 ∂ν r r 2 ∂r ∂r c ⇒ ν = − 1 + c2 r
Ans:
=0
(for spherical coordinate)
Given ν = 0 at r = ∞ ⇒ c2 = 0 ⇒ ν = − When ν = ν a at r = a, ⇒ ν a = −
c1 r
c1 a
And c1 = − aν a aν ∴ ν= a. r
Q.60
What length of transmission line should be used at 500 MHz and how should it be terminated for use as a parallel resonant circuit. (i) (ii) series resonant circuit. (6)
Ans:We have seen that for line to act as parallel circuit 3 × 1010 λ= = 60cm 500 ×10 6 λ 60 = = 15cm λ 4 4 Then for short circuit line (odd multiple of ) short circuit 3λ 4 = 45cm 4 And for open circuited line the length is (even multiple of
λ 4
)
2λ 2 × 60 = = 30cm 4 4 open circuit 4λ = 60cm 4
Q.61
In a rectangular waveguide for which a = 1.5 cm, b = 0.8 cm, σ = 0, µ = µ o and ∈= 4 ∈o and πx 3πy 11 H x = 2 sin cos Sin π × 10 t − β z A m a b Find (i) the mode of operation (ii) the cut off frequency (iii) phase constant
(
)
52
AE14
ELECTROMAGNETICS AND RADIATION (iv) (v)
Ans: (i)
TM 13
(ii)
f cmn =
or
TE13
µ ′ m2 2
a
2
+
n2 , b2
µ′ =
f β = W µ ∈ 1 − c f
1
µ∈
f =
100 = 50GH Z 2
γ = jβ = j1718.81 / m 2
Q.62
c 2
2
Where W = 2πf = π × 1011 or
(v)
=
2
f W ∈r = 1 − c C f β = 1718.81 rad/m
(iv)
(8)
f c13 = 28.57GH Z
Or (iii)
the propagation constant the intrinsic wave impedance
ηTM
13
2
f 377 28.57 = η 1 1 − c = 1− = 154.7Ω ∈r 50 f
What is a linear antenna? Find out the expression for effective length of linear antenna when the current is distributed along its length and (i) (ii) there is sinusoidal distribution of current. Also find the total power radiated. (8)
Ans: (i) Refer Text Book – II, Chapter – 18, 21. (ii) Refer Text Book – I, Chapter – 11. Q.63
A small dipole antenna is carrying a uniform r.m.s current of 10 amperes. Its r.m.s. electric field at a distance ‘r’ metre in a direction making an angle ‘ θ ’ with the conductor is given by 200 E= Sin θ V m r Find the total power radiated and radiation resistance. (8)
Ans:Since E = =
(
60πlo dl sin θ cos ω t − r c λr
200 sin θ r
E2 1 200 Pav = EH = = sin θ 120π 120π r
2
53
)
AE14
ELECTROMAGNETICS AND RADIATION =
1000 2 sin θ 3πr 2 π
And total power, P = ∫ Pav .2πr 2 sin θdθ 0
1000 3 sin θ .2πr 2 dθ 3πr 2 = 888.8ω
=
Q.64
What is skip distance? Find the expression for skip distance and maximum usable frequency considering flat surface and curved surface of earth. (10)
Ans:Refer text Book – II Chapter – 19 Flat Earth => f c = f cos i Dskip = 2h
f2 −1 2 fc
D 2h Curved Earth tan i =
D = 2 Rθ BT R sin θ tan i = = AT h + R − R cos θ h Since >a, at the centre of the solenoid, H = nIa z θ1
a
θ2 P
(8)
z
λ Ans:Since solenoid consists of circular loops, so we can apply result of Idλa 2 dH 2 = 3 2 a2 + z2 2
[
=
]
za 2 ndz
[
2 a2 + z2
]
3
2
r The contribution to H at P by element of solenoid of length d z . a Where dλ = ndz = N dz also tan θ = λ z
( ) (
− z2 + a2 dz = − a cos ec dθ = a2 Hence, − nI dH 2 = sin θdθ 2 θ − nI 2 H2 = sin θdθ 2 θ∫1 2
)
3
2
sin θdθ
Thus r nI H = (cos θ 2 − cos θ1 )a z 2 Substituting n = N
λ
58
AE14
ELECTROMAGNETICS AND RADIATION r NI H= (cos θ 2 − cos θ1 )a z 2λ However at the centre cos θ 2 = − cos θ1 and
If λ >> a or θ 2 = 00 θ1 = 180 0 r NI ∴ H = nI a z = az
λ
Q.70
Write down Maxwell’s equations in integral as well as differential forms for time-varying fields. (8)
Ans:Maxwel’s Eqauations for time varying fields: DIFFERENTIAL FORMS INTEGRAL FORMS 1 1 ρ E.da = ρdV I. ∇.e = ∫ Σ0 ε0 ∫ II. ∇.B = 0 III. ∇ × E = −
∂B ∂t ∂E ∂t
IV. ∇ × B = µ 0 J + µ 0 Σ 0
Q.71
∫
B.dA = 0
∫
E.dl = − ∫
∫
µ ∂E B.dl = ∫ µ 0 J + ε 0 0 .dA . ∂t
∂B .dA ∂t
In a medium characterized by the parameters σ =0, µ0 and ∈0 , and an electric field given by
(
)
E = 20 sin 10 8 t − β z a z V/m, calculate β and H. r dt Ans: ∇.E = y = 0 dy From Faraday’s law r dH ∇.E = − µ dt −1 H= ∫ (∇ × E )dt
(as σ = 0 ∇.Dρv = 0 )
µ
But
ax
ay
az
r d ∇× E = dx 0
d dy 0
d dz Ez
=
dE z d ax − Ez a y dy dz
= 0 − 20 β cos(108 t − β z )a y
59
(8)
AE14
ELECTROMAGNETICS AND RADIATION r 20 β 8 H= ∫ cos(10 t − βz )dt a y
µ
=
β × 20 sin(108 t − βz )a y µ108
Also r 1 r E = ∫ (∇ × H )dt ∈ r dH y β 2 × 20 ∇× H = − ax = cos(108 t − β z )a x 8 µ10 dz 2 r β 20 E= × cos(108 t − βz )dt a x 8 ∫ ∈ µ10 =
β 2 × 20 sin(108 t − βz )a x 8 8 ∈ µ10 × 10
Comparing for E we get 20 β 2 = 20 µ ∈ ×1016 β 2 = µ ∈ ×1016
β = ±108 µ ∈ = ± 108 µ 0 ∈0 108 C 108 =± 3 × 108 =±1 3 r 1 20 z sin 108 t ± a y H= × −7 8 3 4π × 10 × 10 3 1 z = sin 108 t ± a y . 6π 3 =±
Q.72
Obtain an expression for the propagation constant in good conductors. Explain skin effect. (8) Ans:Helmholtz equation ∇2 E − y 2 E = 0 Let y′ = α + jβ (propagation constant) Since y 2 = jωµ (σ + jω ∈) Where α = attenuation constant Β = phase shift constant. Uniform wave travelling in x direction ∂2E = y2E 2 ∂x
60
AE14
ELECTROMAGNETICS AND RADIATION E ( x) = E0 e − yz Time varying form r E ( x, t ) = Re E0e + jωt − yx
{
=e
− ax
Re{E0 e
Q α + jβ =
}
j (ωt − β x )
}
jωµ (σ + jω ∈)
α 2 + β 2 + j 2αβ = jωµ (σ + jω ∈) σ = jωµ × jω ∈ 1 + jω ∈ σ = − ω 2 µ ∈ 1 + jω ∈ Equating real and imaginary part. α 2 − β 2 = −ω 2 µ ∈ 2αβ = jωµσ . Solving for α and β, we get
α =ω
µ ∈
2 σ 1+ − 1 2 ω ∈
µ ∈
2 σ 1+ β =ω + 1 2 ω ∈ SKIN EFFECT: Skin depth ∂ is defined as the distance it takes to reduce the amplitude to 1/e about 3.7% 2 1 1 ∂= = ~ α k ωµσ So, in good conductor, the depth of penetration decreases with frequency low. In term of th wave length ∂ = λ i.e. about 1 2π σ of a wave length. So a wave gets attenuated even before one cycle. This phenomena of conductor is known as skin effect. So what is a good conductor at low or audio frequencies may become poor at high frequencies.
Q.73
A lossy dielectric has an intrinsic impedance of 200 ∠ 30° Ω at a particular frequency. If at that frequency, the plane wave propagating through the dielectric has the magnetic field component 1 H = 10e − αx cos ωt − x a y A/m, Find E and α. Determine the skin depth and wave 2 polarization. (8) r Ans:Wave travels along a x so that K = a x ; H = a y r r ∴ − E = K × H = ax × a y = az r E = −a z
61
AE14
ELECTROMAGNETICS AND RADIATION Also Ho = 10 so E
Ho
= η = 200∠30 = 200e jπ / 6
Eo = 2000 e jπ / 6 r r ∴ E = Re 2 × 103 e jπ / 6 .e − yx .e jωt .E
[
]
r r So except for amplitude and phase difference E and H have same form r x π E = −2 ×10 3 e − ax cos ωt − + a z 2 6 Knowing that β = 1 / 2 , so α will be 2 σ 1 + −1 α ωt = 2 β 1 + σ + 1 ωt
Where
1
2
σ = tan 60 = 3 ω∈ 1
1 α 2 − 1 So = = β 2 + 1 3 1 2 1
α= × δ= Q.74
α
2
1 1 = 3 2 3
=2 3m.
What are standing waves? How do they arise? Discuss their characteristics.
(8)
Ans:When a line is terminated in an iimpedance other than Z c , there is a mismatch and consequent reflection of voltage or current from the load end. If there is a mismatch at sending end also, the wave is re-reflected. Thus, there will be multiple reflections and standing waves on the line. On a reflected wave, the distance between successive maxima or minima is λ . Similarly, 2 the distance between a minimum and maximum is λ . 4 1. The maximum amplitude is Vmax = Ae − ax + Be ax Where βx = nπ where n = 0, ± 1, ± 2 etc. 2. The maximum amplitude is Vmin = Ae − ax − Be + ax Where β = (2n − 1) π
3.
where n = 0, ± 1, ± 2 etc. 2 The distance between any two successive maxima or minima is nπ nλ β x = nπ ; x = = 2 β
62
AE14 Q.75
ELECTROMAGNETICS AND RADIATION A 30-m long lossless transmission line with Z0=50Ω operating at 2 MHz is terminated with a load ZL=60+j40 Ω. If the velocity of propagation on the line is 0.6 times that of light, find the following without using Smith-chart: (i) The reflection coefficient Γ (ii) The standing wave ratio S (iii) The input impedance (8)
Ans: (i)
(ii) (iii)
Reflection coefficient Γ 60 + j 40 − 50 Γ= 60 + j 40 + 50 10 + j 40 = 110 + j 40 = 0.3523 ∠560 1+ Γ VSWR = = 2.088 1− Γ Z + jZ 0 tan( β d ) Z in = Z 0 L Z 0 + jZ L tan( β d )
βd =
ωl 2π (2 × 106 )(30) = = 120 0 8 0.6 × (3 ×10 ) µ
60 + j 40 + j 50 tan(120) Z in = 50 60 + j[60 + j 40] tan(120) = 23.97 + j1.35 .
Q.76
What is a cavity resonator? Derive an expression for the frequency of oscillation of a rectangular cavity resonator. (8)
Ans:Cavity Resonator:A cavity resonator is a hollow inductor blocked at both ends and along which are electromagnetic wave can be supported. It can be viewed as a wave guide short circuited at both ends. The cavity has interior surfaces which reflects a wave of a specific frequency. Where a wave that is resonant with the cavity enters, it bounces back and forth within the cavity, with low loss. As more wave energy enters the cavity, it combines with and reinforces the standing wave, increasing its intensity. Q.77
In a rectangular waveguide for which a=1.5cm, b=0.8cm, σ = 0, µ = µ 0 ,
(
)
πx 3πy 11 and ∈= 4∈0 , H x = 2 sin cos sin πx10 t − β z A/m a b Determine (i) The mode of operation (ii) The cutoff frequency (iii) The phase constant β (iv) The intrinsic wave impedance η
63
(8)
AE14
ELECTROMAGNETICS AND RADIATION Ans: (i)
(ii)
The wave guide is operating at TM 13 or TE13 mode. fcmm = =
µ′ m2 a2
2 C 4
+
n2 b2
1
9
+
[1.5 ×10 ] [0.8 ×10 ] −2 2
−2 2
= 28.57 GH z (iii)
fc β = ω µ ∈ 1 − f
ω ∈r
=
C
2
fc 1 − f
2
28.57 1− β= 8 3 × 10 50 = 1718.81 rad/m
π × 1011 (2)
(iv)
ηTM
13
fc = η ′ 1 − f
2
2
377 28.57 1− ∈r 50 = 154.7Ω .
2
=
Q.78
Explain the principle of a phased array, and write a note on the log-periodic dipole array.
(8) Ans:PHASED ARRAY:Phased Array is designed with more energy radiated in some particular directions and less in other directions, i.e. the radiation pattern be concentrated in the direction of interests. A phased array is used to obtain greater directivity that cab be obtained with a single antenna element. A phased array is a group of radiating elements arranged so as to produce some particular radiation characteristics. It is practical and convenient that the array consists of identical elements, but this is not fundamentally required. LOG PERIODIC DIPOLE ARRAY: The log periodic dipole array (LPDA) is one antenna that almost everyone over 40 years old has seen. They were used for years as TV antennas. The chief advantage of an LPDA is that it is frequency-independent. Its input impedance and gain remain more or less constant over its operating bandwidth, which can be very large. Practical designs can have a bandwidth of an octave or more. Although an LPDA contains a large number of dipole elements, only 2 or 3 are active at any given frequency in the operating range. The electromagnetic fields produced by these active elements end of the array. The radiation in the opposite direction is typically 15 – 20 dB 64
AE14
ELECTROMAGNETICS AND RADIATION below the maximum. The ratio of maximum forward to minimum rearward radiation is called the Front-to-Back (FB) ration and is normally measured in dB.
The log periodic antenna is characterized by three interrelated parameters, α, σ , and τ as well as the minimum and maximum operating frequencies, f MIN and f MAX . The diagram below shows the relationship between these parameters.
Q.79
Unlike many antenna arrays, the design equations for the LPDA are relatively simple to work with. π A magnetic field strength of 5µA/m is required at a point θ = , 2 km from an antenna in 2 air. Neglecting ohmic loss, how much power must the antenna transmit if it is
65
AE14
ELECTROMAGNETICS AND RADIATION λ 25
(i)
A Hertizian dipole of
(ii) (iii)
A half-wave dipole A quarter-wave monopole.
Ans: (i)
For a Hertizian dipole I β dl sin θ H ϕs = 0 where ∂l = λ 25 4πr 2π λ 2π β dl = . = λ 25 25 Thus from (1) We get I 0 = 0.5 A
(8)
------------------ (1)
2
2 2 dl 2 40π (0.5) Prad = 40π I 0 = (25) 2 λ = 158 mω 2
(ii)
For a λ
2
dipole
π I 0 cos . cos θ 2 H ϕs = 2πr sin θ ⇒ I 0 = 20π mA
1 2 1× (20π ) 2 I 0 Rrad = .10 − 6 2 2 = 144 mω .
Pav =
(iii)
Q.80
For a λ
monopole 4 I 0 = 20π mA As in part (b) 1 2 1 Prad = I 0 Rrad = (20π ) 2 × 10 −6 × 36.56 2 2 = 72 mω
Explain Poisson’s and Laplace’s equations with suitable applications.
Ans:POISSON’S AND LAPLACE’S EQUATIONS: Point form of Gauss Law ∇.D = ρ v Relation between D & E D =∈ E Relation between E & V E = −∇V Hence ∇.D = ∇.(∈ E ) = −∇.(∈ ∇ V ) = ρ v
66
(8)
AE14
ELECTROMAGNETICS AND RADIATION ∇.∇.V = −
ρv
∈ For a homogenous region in which ∈ is a constant. The last equation is known as poisson’s equation. In cartessan co-ordinate system ∂A ∂A ∂A ∇. A = x + y + z ∂x ∂y ∂z ∂v ∂v ∂v ∇.V = a x + a y + a z ∂x ∂y ∂z And therefore ∂ 2v ∂ 2 v ∂ 2v ∇.∇ V = 2 + 2 + 2 ∂x ∂y ∂z Usually the operation ∇ . ∇ is ∇ 2V , hence ∂ 2 v ∂ 2v ∂ 2v ∇ 2V = 2 + 2 + 2 ∂x ∂y ∂z − ρv ∇ 2V = ∈ If ρ v = 0 ∇ 2V = 0 Which is the Laplace’s equation. The ∇ 2V is called the laplacian of V.
Q.81
The surface of the photoconductor in a xerographic copying machine is charged uniformly with surface charge density ρs as shown in figure. When the light from the document to be copied is focused on the photoconductor, the charges on the lower surface combine with those on the upper surface to neutralize each other. The image is developed by pouring a charged black powder over the surface of the photopowder, which is later transferred to paper and melted to form a permanent image. Determine the electric field below and above the surface of a photoconductor. (8) x
E1
air
d a
light
∈1
ρs
+++++
photo conductor
E2
∈2
+++++ recombination
–––––
Ans:Since ρ v = 0 in this case, we apply Laplace’s equation. Also the potential depends only on x. Thus ∂ 2V 2 ∇ V = 2 =0 ∂x Integrating twice gives V = Ax + B Let the potential above and below be V1 and V2 , respectively. V1 = A1 x + B1 , x > a 67
AE14
ELECTROMAGNETICS AND RADIATION
V2 = A2 x + B2 , x < a ……………….1 The boundary condition at the grounded electrodes are V1 ( x = d ) = 0 V2 ( x = 0) = 0 ……………….2 At the surface of the photoconductor, V1 ( x = a) = V2 ( x = a ) D1n − D2 n = ρ s x = a ……………….3 We use the four conditions in eqs. (2) and (3) to determine the four unknown constants A1 , A2 , B1 and B2 . From eqs. (1) and (2), 0 = A1d + B1 → B1 = − A1d 0 = 0 + B2 → B2 = 0 From eqs. (1) and (2), A1a + B1 = A2 a To apply eq. (2), recall that D = εE = −ε∇V so that dV dV ρ s = D1n − D2 n = ε 1 E1n − ε 2 E2 n = −ε 1 1 + ε 2 2 dx dx Or ρ s = −ε 1 A1 + ε 2 A2 Solving for A1 and A2 in eqs. (6.2.4) to (6.2.6), we obtain ρ s ax E1 = − A1a x = ε d ε ε 1 1 + 2 − 2 ε1 a ε1 d − ρ s − 1 a x a E2 = − A2 a x = ε d ε ε 1 1 + 2 − 2 ε 1 a ε1
Q.82
What is Gauss law? How gauss law is applicable to point charge and infinite line charge. (4) Ans:The generalization of Faraday’s experiment lead to the following statement known as Gauss’s Law “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface.” For a given fig if the total charge is φ, then φ of electric flux pass through the enclosing surface. Let at point P, an is a unit vector normal to differential area ∆s located at point P. Then ∆s = ∆s an ∆ϕ = flux crossing ∆s = Ds noun ∆s = Ds cos θ ∆s And the total flux is given by
ϕ =
∫ dϕ = ∫
s
Ds .ds
Thus mathematical formulation of Gauss’s law is ϕ = Q = φs Ds.ds
68
AE14
ELECTROMAGNETICS AND RADIATION
For point charge:
ϕ = ∫ Ds.ds = ∫ s
spn
2π π
Dsds = Ds ∫ ∫ r 2 sin θ dθdϕ 0 0
2
= 4 πr Ds. Hence Q Q Ds = . & E= ar 2 4πr 4πε 0 r 2 For infinite line charge: Q=∫ Ds.ds = Ds ∫ ds + 0 ∫ ds + 0
∫ ds
cyc
sides
top
bottom
L 2π
= Ds ∫ ∫ ρdϕdz = Ds 2πρL 0 0
Ds =
Q
2πρL Hence Q ρ ×L E= υ= L υ 2πρε 0 L 2πρε 0 L =
Q.83
ρL υ. 2πε 0 ρ
The spherical region 0 ≤ r ≤ 3 contains uniform volume charge density of ρ v = 2 C / m 3 and ρ v = 1 C m 3 for 5 ≤ r ≤ 6 . Use law to find D r for (i) r ≤ 3 (ii) 3 ≤ r ≤ 5 (iii) 5 ≤ r ≤ 6 (iv) r>>6
Ans: (i)
For r ≤ 3 r π 2π
r π 2π
0 0 0
0 0 0
ψ = ∫ ∫ ∫ ρ v dv = ∫ ∫ ∫ 2r 2 sin θdθ dϕ
69
(8)
AE14
ELECTROMAGNETICS AND RADIATION =
(ii)
(iii)
(iv)
2r 3 × 4π 3
2r 3 × 4π ψ 2r 3 Dr = = = ar 2 area 4πr 3 For 3 ≤ r ≤ 5 Will be same as (i) 2r Dr = ar 3 For 5 ≤ r ≤ 6 r π 2π 2r 3 ψ= × 4π + ∫ ∫ ∫ r 2 sin θdθ dϕ 3 0 0 0 2r 3 r3 × 4π + × 4π = 3 3 3 4πr ψ D= = ar = r ar area 4πr 2 For r >> 6 it will same as for r = 5 ≤ r ≤ 6 Thus Dr = r.ar .
Q.84
Write Laplaces equation in Cartesian, Cylindrical, Spherical coordinates.
(4)
Ans:CARTESIAN ∂ 2V ∂ 2V ∂ 2V ∇ 2V = 2 + 2 + 2 ∂x ∂y ∂z CYLINDRICAL 1 ∂ ∂V 1 ∂ 2V ∂ 2V ∇ 2V = ρ + + ρ ∂ρ ∂ρ ρ 2 ∂ϕ 2 ∂z 2 SPHERICAL 1 ∂ 2 ∂V 1 ∂ ∂V 1 ∂ 2V 2 ∇V= 2 . r + sin θ + r ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin θ ∂ϕ 2 Q.85
Define Biot Savert law. Calculate the magnetic field of line current along a thin straight wire of infinite length. (6)
Ans:Biot-Savert Law may be written as
I dL × a12 4π .d 2 Where d = distance of point where it is to be find out from element dl. a12 = unit vector from point 1 to point 2. I = current in the filamentary conductor. FOR WIRE OF INFINITE LENGTH: dH =
70
AE14
ELECTROMAGNETICS AND RADIATION Suppose a wire of infinite length is kept on z axis carries a current I. Then any point 1(0,0,z) on the z axis at which a differential length. dL = d z a z is located. The point at which H has to be calculated is 2( ρ , ϕ , τ ) R12 = r2 − r1 r = ρυ + z a z − z a z r = ρυ + ( z − z )a z
d = R12 = ρ 2 + (Z − z )
2
∴ dH =
[
∞
∴ H=
[
Idz a z × ρυ + ( Z − z )a z 2 4π ρ 2 + (Z − z ) 3 2
]
]
Idz ( ρ aϕ )
∫ 4π [ρ + (Z − z ) ]
2 3
2
−∞
2 ∞
Iρ aϕ 1 Z−z − 2 = 2 4π ρ ρ + ( Z − z ) −∞ I = aϕ 2∏ ρ In general the magnetic field intensity due to infinitely long filamentary conductor is given I by H = a1 × a12 2πd Where a1 , is a unit vector in the direction of current, point 2 is that point at which magnetic field intensity is derived, point 1 is the foot of perpendicular from point 2 on the filamentary conductor and d is the distance between the joints 1 & 2.
(
Q.86
)
Find the magnetic flux density and field intensity at a point P due to a straight conductor carrying a current I as (10)
Ans:Magnetic field due to length element dl is
I dL × ar 4πR 2 I dL × sin(α )aϕ = 4πR 2 Where Rdα = dl sin α And R = h sin α IR dα × sin α aϕ Idα sin α dH = = aϕ 2 4πR 4πR Idα = aϕ 4π h sin α dH =
(
)
71
AE14
ELECTROMAGNETICS AND RADIATION H=
I
α2
4πh α∫1 I
sin α dα aϕ
[− cos α ]αα
aϕ 4πh I [cos α1 − cos α 2 ]aϕ = 4πh
=
Q.87
2
1
Given that E = 50πe j(ωt −β z ) (u x ) H = H m e j(ωt −βz ) (u y ) in free space where ω = 109 .
Evaluate H m and β (β > 0) Ans:Given ω = 109
∴ β=
(8)
ω c
=
109 = 10 3 3 × 108
Now r 1 r r E = ∫ ∇ × Hdt Q ∠ = 0
ε
v r −∂ ∂ ∇ × H = 2 H y ax + az Hy ∂ ∂x = β H m e j (ωt − βz ) a x r 1 E = ∫ β H m e j (ωt − βz )dt a x
ε β H m j (ωt − βz ) = e ax ωε r
Comparing with E we get βH m = 50π
ωε
50π × 3 × ωε β 10 = 15π ×109 × 8.854 × 10 −12 = 416.3 × 10 −3 = 0.4163 A/m.
Hm =
Q.88
50πωε
=
Show that E and H fields constitute a wave travelling in Z-direction. Verify that the wave (8) speed and E/H depend only on the properties of free space.
Ans:Relation between H & E for a wave propagation in z direction can be written as E x ( z , t ) = E X 0 cos(ωt − k0 z ) --------------------- (1) Applying maxwells equation H can be easily found out as
H y ( z, t ) = E X 0
ε0 cos(ωt − k0 z ) µ0
--------------------- (2)
72
AE14
ELECTROMAGNETICS AND RADIATION Therefore from above relations we find that for an X directed E field that propagates the wave in z direction, the direction of H field comes out to be y directed. Thus a uniform plane wave is transverse in nature. Taking ratio of (1) & (2) we get µ0 Ex = ---------------------- (3) Hy ε0 Also k0 = ω
v =ω v=
and λ = 2λ
c
k0
β = ω µ 0ε 0
β 1
µ 0ε 0
From above relation we find that wave speed and E/H depends only on the properties of free space,
Q.89
Define polarization of waves, linear polarization, elliptical polarization, circular polarization. (4)
Ans:Polarization refers to time varying behaviour of electric field strength vector at some fixed point in space. LINEAR POLARIZATION: For a wave to have linear polarization, the time phase difference between two component must be δ = δ y − δ y = nπ Where n = 0,1,2,3,4---------A linearly polarized wave can be resolved into a light hand circularly polarized wave and left hand circularly polarized wave of equal amplitude. ELLIPTICAL POLARIZATION: E0 = a x A + j a y B & E (o, t ) = A cos ωt a x − B sin ωt a y
E x = A cos ωt 2
E y = − B sin ωt
2
E Ex + y2 = 1 ∴ END points E (o, t ) traces out an Ellipse. 2 A B CIRCULAR POLARIZATION: If y component leads x by 90 0 Ea = amplitude
∴
(
)
(
)
E0 = a x + j a y . Ea ⇒ E (o, t ) = cos ωt a x − sin ωt a y Ea
E x = Ea cos ωt 2 2 2 E x + E y = Ea E y = − Ea sin ωt End points of E (o, t ) traces out a circle of radius Ea .
Q.90
is required at a point on θ = π , 2 Km from m 2 antenna in air. Neglecting ohmic loss how much power must antenna transmit if it is
A magnetic field strength of 5µ A
73
an
AE14
ELECTROMAGNETICS AND RADIATION (i)
a hertzian dipole of length λ
(ii)
A half wave dipole.
25
.
(4+4)
Ans: (i) For a Hertzian dipole I β dl sin θ H ϕs = 0 4πr Where dl = λ 25 2π λ 2π βdl = × = λ 25 25 I . 2π 25(1) I 0 5 × 10 −6 = 0 = 5 4π 2 × 103 10 I 0 = 0 .5 A
(
)
2
(ii)
Q.91
2 2 dl 2 40π (0.5) Prad = 40π 2 I 0 = (25) 2 λ = 158 mw. For a λ/2 dipole π I 0 cos . cos θ 2 H ϕs = 2πr sin θ I 0 .1 5 × 10 − 6 = 2π (2 × 103 )(1) I 0 = 20π mA 1 2 1 Pav = I 0 Rrad = (20π ) 2 × 10 − 6 2 2 = 144 mw.
Find the characteristic impedance of lossless transmission line having R=5Ω, L=40H and C=10F having frequency of 10Hz. (4)
Ans:A transmission line is said to be lossless if both its conductor & dielectric loss are zero, R = 0 and G = 0 L ∴ Characteristic impedance = C 40 = = 4 10 = 2 Ω. Q.92
What is standing wave ratio? Calculate reflection coefficient having SWR of 1.5.
(4)
Ans:Standing Wave ratio: The ratio of maximum to minimum voltages along a finite terminated line is called standing wave ratio. 74
AE14
ELECTROMAGNETICS AND RADIATION
SWR =
Vmax V min
=
1 + ΓL 1 − ΓL
∴ ΓL = reflection coefficient is SWR − 1 = SWR + 1 1 .5 − 1 .5 1 = = = . 3 1.5 + 1 1.5 Q.93
Define cut-off wavelength for a rectangular wave guide. A rectangular wave guide measures 3 x 4.5 cm internally and has a 10GHz signal propagated in it. Calculate the cutoff wavelength, the guide wavelength and characteristic wave impedance for TE10 mode. (8)
Ans:For TE10 mode f ′ = 10 GH z λg = 0.8757λc for TE10 mode
c 3 × 108 = × 100 f 10 × 109 = 3 cm.
λ 0=
2
2
λg λ = 1 + g λ0 λc = 1+ (0.8757) 2 = 1.767
λg = 1.329 & λg = 1.329 × λ0 = 4.00 cm λ0 λg 4 = = 4.571 cm λc = 0.8757
ηTE10 Q.94
0.8757 4 = 120 × π × = 160π . 3
Explain Hertizan dipole. Show time variation of current and charge in Hertizan dipole.
(8)
Ans:Hertzian dipole – By Hertzian dipole, we mean an infinite current element idl. Although such a current element does not exist in real life, it serves as building block from which the field of a practical antenna can be calculated by integration. Consider a hertzian dipole as in figure. Magnetic vector potential at point P, due to dipole is µIdl A= a z 4πr Where I is given by wr [ I ] = I 0 cos wt − a = I 0 cos(wt − β r ) 75
AE14
ELECTROMAGNETICS AND RADIATION We find the E field using ∇ × H as 1 1 ηI dl Exs = 0 cos θ 2 − 3 e − jβr 2π βr r Eys =
jβ 1 j ηI 0 dl sin θ + 2 − 3 e − jβr 4π βr r r
Ezs = 0 Power dissipated in fictitions Resistance Rrad . 1 2 Prad = I 0 Rrad 2 2
Where Rrad
Q.95
dl = 80π . λ 2
Define radiation resistance and directivity. Calculate the radiation resistance of an antenna having wavelength λ = 5 and length 25cm. (8)
Ans: Rrad
dl = 80π λ
2
2
25 = 80π 2 5 = 2000π 2
2
Directivity – The directivity D of an antenna is the ratio of maximum radiation intensity to the average radiation intensity.
Q.96
Write short notes (i) Space wave propagation (ii) Skip distance (iii) Ground wave propagation (iv) Antenna Array
(16)
Ans: (i) SPACE (DIRECT) WAVE PROPAGATION Space Waves, also known as direct waves, are radio waves that travel directly from the transmitting antenna to the receiving antenna. In order for this to occur, the two antennas must be able to “see” each other; that is there must be a line of sight path between them. The diagram on the next page shows a typical line of sight. The maximum line of sight distance between two antennas depends on the height of each antenna. If the heights are measured in feet, the maximum line of sight, in miles, is given by; d = 2 ht + 2 h r Because a typical transmission path is filled with buildings, hills and other obstacles, it is possible for radio waves to be reflected by these obstacles, resulting in radio waves that arrive at the receive antenna from several different directions. Because the length of 76
AE14
ELECTROMAGNETICS AND RADIATION
(ii)
(iii)
each path is different, the waves will not arrive in phase. They may reinforce each other or cancel each other, depending on the phase differences. This situation is known as multipath propagation. It can cause major distortion to certain types of signals. Ghost images seen on broadcast TV signals are the result of multipath-one picture arrives slightly later than the other and is shifted in position on the screen. Multipath is very troublesome for mobile communications. When the transmitter and/or receiver are in motion, the path lengths are continuously changing and the signal fluctuates wildly in amplitude. For this reason, NBFM is used almost exclusively for mobile communications. Amplitude variations caused by multipath that make AM unreadable are eliminated by the limiter stage in an NBFM receiver. An interesting example of direct communications is satellite communications. If a satellite is placed in an orbit 22,000 miles above the equator, it appears to stand still in the sky, as viewed from the ground. A high gain antenna can be pointed at the satellite to transmit signals to it. The satellite is used as a relay station, from which approximately ¼ of the earth’s surface is visible. The satellite receives signals from the ground at one frequency, known as the uplink frequency, translates this frequency to a different frequency, known as the downlink frequency, and retransmits the signal. Because two frequencies are used, the reception and transmission can happen simultaneously. A satellite operating in this way is known as a transponder. The satellite has a tremendous line of sight from its vantage point in space and many ground stations can communicate through a single satellite. SKIP DISTANCE/SKIP ZONE – The SKIP DISTANCE is the distance from the transmitter to the point where the sky wave is first returned to Earth. The size of the skip distance depends on the frequency of the wave, the angle of incidence, and the degree of ionization present. The SKIP ZONE is a zone of silence between the point where the ground wave becomes too weak for reception and the point where the sky wave is first returned to Earth. The size of the skip zone depends on the extent of the ground wave coverage and the skip distance. When the ground wave coverage is great enough or the skip distance is short enough that no zone of silence occurs, there is no skip zone. Occasionally, the first sky wave will return to Earth within the range of the ground wave. If the sky wave and ground wave are nearly of equal intensity, the sky wave alternately reinforces and cancels the ground wave, causing severe fading. This is caused by the phase difference between the two waves, a result of the longer path travelled by the sky wave. PROPAGATION PATH: The path that a refracted wave follows to the receiver depends on the angle at which the wave strikes the ionosphere. You should remember, however, that the rf energy radiated by a transmitting antenna spreads out with distance. The energy therefore strikes the ionosphere at many different angles rather than a single angle. After the rf energy of a given frequency enters an ionospheric region, the paths that this energy might follow are many. It may reach the receiving antenna via two or more paths through a single layer. GROUND WAVE PROPAGATION Ground Waves are radio waves that follow the curvature of the earth. Ground waves are always vertically polarized, because a horizontally polarized ground wave would be shorted out by the conductivity of the ground. Because ground waves are actually in contact with the ground, they are greatly affected by the ground’s properties. Because ground is not a perfect electrical conductor, ground waves are attenuated as they follow the earth’s surface. This effect is more pronounced at higher frequencies, limiting the
77
AE14
ELECTROMAGNETICS AND RADIATION
(iv)
usefulness of ground wave propagation to frequencies below 2 MHz. Ground waves will propagate long distances over sea water, due to its high conductivity. Ground waves are used primarily for local AM broadcasting and communications with submarines. Submarine communications take place at frequencies well below 10KHz, which can penetrate sea water (remember the skin effect?) and which are propagated globally by ground waves. Lower frequencies (between 30 and 3,000 KHz) have the property of following the curvature of the earth via groundwave propagation in the majority of occurrences. In this mode the radio wave propagates by interacting with the semi-conductive surface of the earth. The wave “clings” to the surface and thus follows the curvature of the earth. Vertical polarization is used to alleviate short circuiting the electric field through the conductivity of the ground. Since the ground is not a perfect electrical conductor, ground waves are attenuated rapidly as they follow the earth’s surface. Attenuation is proportional to the frequency making this mode mainly useful for LF and VLF frequencies. Today LF and VLF are mostly used for time signals, and for military communications, especially with ships and submarines. Early commercial and professional radio services relied exclusively on long wave, low frequencies and ground-wave propagation. To prevent interference with these service, amateur and experimental transmitters were restricted to the higher (HF) frequencies, felt to be useless since their ground-wave range was limited. Upon discovery of the other propagation modes possible at medium wave and short wave frequencies, the advantages of HF for commercial and military purposes became apparent. Amateur experimentation was then confined only to authorized frequency segments in the range. ANTENNA ARRAY An antenna array is an antenna that is composed of more than one conductor. There are two types of antenna arrays: Driven arrays – all elements in the antenna are fed RF from the transmitter. Parasitic arrays – only one element is connected to the transmitter. The other elements are coupled to the driven element through the electric fields and magnetic fields that exist in the near field region of the driven element. There are many types of driven arrays. The four most common types are: COLLINEAR ARRAY The collinear array consists of λ/2 dipoles oriented end-to-end. The center dipole is fed by the transmitter and sections of shorted transmission line known as phasing lines connect the ends of the dipoles. BROADSIDE ARRAY A broadside array consists of an array of dipoles mounted one above another as shown below. Each dipole has its own feed line and the lengths of all feed lines are equal so that the currents in all the dipoles are in phase. LOG PERIODIC DIPOLE ARRAY The lop periodic dipole array (LPDA) is one antenna that almost everyone over 40 years old has seen. They were used for years as TV antennas. The chief advantage of an LPDA is that it is frequency-independent. Its input impedance and gain remain more or less constant over its operating bandwidth, which can be very large. Practical designs can have a bandwidth of an octave or more. YAGI-UDA ARRAY (YAGI)
78
AE14
ELECTROMAGNETICS AND RADIATION The Agi-Uda array, named after the two Japanese physicists who invented it, is the most common antenna array in use today. In contrast to the other antenna arrays that we have already looked at, the Yagi has only a single element that is connected to the transmitter, called the driver or driven element. The remaining elements are coupled to the driven element through its electromagnetic field. The other elements absorb some of the electromagnetic energy radiated by the driver and re-radiate it. The fields of the driver and the remaining elements sum up to produce a unidirectional pattern. The diagram below shows the layout of elements in a typical Yagi.
Q.97
Prove that energy density stored in an electric field of magnitude E is proportional to E 2 . (8) Ans: Refer text Book 1, Section 4.8 (Page No. 111)
Q.98
A circular ring of radius 'a' carries a uniform charge ρ C m and is placed on xy-plane with L
the axis the same as the z-axis (i) Find E (0,0, h ) (ii) (iii)
What value of h gives maximum value of E ? If the total charge on the ring is Q, find E as ‘a’ tends to 0 .
Ans: (i) Consider the figure:
dl = adφ
R = a ( − a P ) + ha z 1 R R =| R |= a 2 + h 2 2 , a R = R aR R − aa P + ha z = = 3 2 3 R |R| a 2 + h2 2
[
]
[
]
79
(8)
AE14
ELECTROMAGNETICS AND RADIATION 2π
E=
− aa P + ha z PL adφ ∫ 4π ∈0 φ = 0 a 2 + h 2 3 2
[
]
By symmetry, contributions along aP add up to zero. Thus we are left with the z-component.
E (0,0, h) =
(ii)
dE dh
[
2
+ a2
2
3
3
2
0
2
2
] [h 1
dE dh
+ a2
2
2
0
L
0
2
2 3
3
1
v/m. 2
2
=0
]
+ a 2 − 3h 2 = 0 a ∴ a 2 − 2h 2 = 0 or h = ± m 2
∴
2
PL aha z
dϕ = ] ∫ 2 ∈ [h
] [h + a ] (1) − 32 (h)(2h)[h + a ] Pa = 2E [h + a ] 4π ∈0 h 2 + a 2
For maximum E ,
[h
2π
PL aha z
2
2
(iii) Since the charge is uniformly distributed, the line charge density is PL = ∴ E=
Q 2πa
Qh
3 az 4π ∈0 h 2 + a 2 2 Q a z v/m As a → 0 , E = 4π ∈0 h 2
Q.99
[
]
Derive an expression for the magnetic field due to an infinite plane sheet of uniform urface current density. (8) 80
AE14
ELECTROMAGNETICS AND RADIATION Ans: Refer text Book 1, Section 8.2 (Page No. 237)
Q.100 A circular loop located on x 2 + y 2 = 9, z = 0 carries a direct current of 10A along aˆ φ . (8)
Determine H at (0, 0, 4) and (0, 0, – 4).
Ans:Consider the circular loop shown in Fig.
Idl × R 4πR 3 dl = ρdφaϕ , R = (0,0, h) − ( x, y,0) = − ρa p + ha z dH =
aρ
aφ
az
dl × R = 0 ρdφ 0 = ρhdφa ρ + P 2 dφa z −ρ 0 h I ( ρhdφa ρ + P 2 dφa z ) = dH ρ a ρ + dH z a z 2 32 4π [ ρ + h ] By symmetry, the contributions along a ρ add up to zero ∴ dH =
2
∴ Hρ = 0 H = ∫ dH z a z =
2π
Iρ 2 dφa z
∫ 4π [ρ 0
∴ H=
[
Iρ 2 a z
2
+ h2
]
3
= 2
Iρ 2 2πa z
[
4π ρ 2 + h 2
]
3
2 ρ 2 + h2 2 I = 10A, ρ = 3, h = 4
10(3) 2 a z ∴ H (0,0,4) = = 0.36a z A/m 3 2[9 + 16] 2 For H(0,0,4), h is released by –h
81
]
3
2
AE14
ELECTROMAGNETICS AND RADIATION ∴ H (0,0,−4) = H (0,0,4) = 0.36a z A/m.
Q.101 State and explain Maxwell’s equation in their Integral and differential forms. Derive the corresponding equations for fields varying harmonically with time. (8) Ans: Refer text Book 1, Section 10.4, 10.5 (Page No. 334 - 337)
Q.102 A conducting bar P can slide freely over two conducting rails as shown in Fig.1. Calculate the induced voltage in the bar (i) If the bar is stationed at y = 8 cm and B = 4 cos 10 6 t aˆ mWb/m 2 . z
(ii) If the bar slides at a velocity v = 20 aˆ y m / s and B = 4 aˆ z mWb/m 2 .
(
)
(iii) If the bar slides at a velocity v = 20 aˆ y m/s and B = 4 cos 10 6 t − y aˆ z mWb/m 2 .
(8)
Ans:Consider the fig. mWb (i) B = 4 cos(10 6 t )a z m2 0.08 0.06 ∂B Vemf = − ∫ .ds = ∫ ∫ 4(10 −3 )(10 6 ) sin(10 6 t )dxdy ∂t y =0 x =0 = 4(103 )(0.08)(0.06) sin(106 t )
Vemf = 19.2 sin(106 t )V (ii) u = 20a y m / s , B = 4a z mWb / m 2
Vemf = ∫ (u + B ).dl =
0
∫ (ua
y
× Ba z ).dxa x
x =l
= − uBl = −20(4)(10 −3 )(0.06) ∴ Vemf = −4.8mV
(
)
(iii) u = 20a y m / s , B = 4 cos 10 6 t − y a z
82
mWb m2
AE14
ELECTROMAGNETICS AND RADIATION Vemf = − ∫
∂B .ds + ∫ (u × B ).dl ∂t
0.06 y
∫ ∫ (4)(10
=
−3
)(10 6 ) sin(10 6 t − y ' )dy ' dx
x =0 0 0
∫ [20a
+
y
]
× (4)(10 −3 ) cos(106 t − y )a z .dxa x
0.06 y
= 240 cos(10 6 t − y ' ) − 80(10 −3 )(0.06) cos(10 6 t − y ) 0
= 240 cos(10 t − y ) − 240 cos 10 6 t − 4.8(10 −3 ) cos(10 6 t − y ) 6
Vemf ~ 240 cos(10 6 t − y ) − 240 cos 10 6 t A+ B A− B sin 2 2 y y = 480 sin 10 6 t − sin V . 2 2
cos A − cos B = −2 sin ∴ Vemf
Q.103 State and prove Poynting theorem. Explain the physical interpretation of each terms in it. (8) Ans: Refer text Book 1, Section 11.3 (Page No. 365 - 369) Q.104 Given a uniform plane wave in air as E i = 40 cos(ωt − β z ) aˆ x + 30 sin (ωt − β z ) aˆ y V m a. Find H i . b. If the wave encounters a perfectly conducting plate normal to the z-axis at z = 0, find the reflected wave E r and H r . c. What are the total E and H fields for z ≤ 0 ? d. Calculate the time-average Poynting vectors for z ≤ 0 and z ≥ 0 .
Ans: Ei = 40 cos(ωt − βz )a x + 30 sin (ωt − β z )a y V
m (i) we may treat the wave as consisting of two waves, Ei1 = 40 cos(ωt − β z )a x , Ei 2 = 30 sin (ωt − βz )a y At atmospheric pressure, air has ε r = 1.0006 ~ 1. Thus air may be regarded as free space. Let H i = H i1 + H i 2
H i1 = H i10 cos(ωt − βz )aH1 40 1 = η 0 120π 3π = ak × a E = a z × a x = a y
H i10 = a H1
Ei10
∴ H i1 =
=
1 cos(ωt − β z )a y 3π
83
(8)
AE14
ELECTROMAGNETICS AND RADIATION Similarly H i 2 = H i 20 sin (ωt − βz )a H 2
aH z
E i 20
30 1 = η 0 120π 4π = ak × a E = a z × a y = − a x
H iz 0 =
=
1 sin(ωt − β z )a x 4π 1 1 sin(ωt − βz )a x + H = H i1 + H i 2 = − cos(ωt − β z )a y 4π 3π (ii) Since the medium 2 is perfectly conducting
∴ Hi2 = −
σ2 >> 1 ⇒ η 2
