Electromagnetic Fields Test - 3 - PDF Flipbook
Electromagnetic Fields Test - 3
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GATE
EEE
Electro
MagneticFields
Test-03Solutions
ELECTROMAGNETIC FIELDS
1. A plane wave is propagating in a material characterized by εr =
2.25, = 1 and = 0. What is the value of in rad/m, when
the electric field is given by:
E� = 10 cos(3 × 1010t − βz) a�x v/m
a) 125
b) 150
c) 175
d) 200
Answer: (b)
Solution:
εr = 2.25, μr = 1
v = velocity = ω
β
v = 1 = 1 . 1
√μϵ �μ0ϵ0 √μrϵr
= 3×108 = 3×108 = 2 × 108m/s
√2.25 1.5
ω = 3 × 1010 rad/sec
β = ω = 3×1010 = 150
v 2×108
2. The magnetic flux density created by an infinitely long
conductor carrying a current I at a radial distance R is
a) μ0I
2πR
b) I
2πR
c) μ0I
2πR3
1
d) 4πR2I
3
Answer: (a)
Solution:
Using Ampere’s law
∮L H��⃗. d���⃗l = Ienc
H × 2πR = I
H = 1
2πR
Magnetic flux density B = μH
∴ B = μ0I
2πR
3. If v, w, q stands for voltage, energy and charge, then v can be
expressed as
a) v = dq
dw
b) v = dw
dq
c) dv = dw
dq
d) dv = dq
dw
Answer: (c)
2
Solution:
If Voltage = v, Charge = q and Energy = w
They are related by dv = dw
dq
= 14.8 × 10−2J = 0.148J ≈ 0.15J
4. A lossless line of length λ/4 and characteristic impedance Z0
transforms a resistive load R into an impedance (Z02/R). When
the line is λ/2 long, the transformed impedance will be
a) (Z02/R)
b) 2(Z02/R)
c) Z0
d) R
Answer: (a)
Solution:
β = 2π
λ
βl = �2λπ� �4λ� = π
2
ZL = R
Zin = Z0�RZ0++jZjR0 tan ππ//22� = Z02
tan R
Zin = Z0�ttaannRZπ0π//22++jjZR0� = Z02
R
For, l = λ , βl = �2λπ� �2l � = π
2
Zin = Z0�RZ0++jZjR0 tan ππ�
tan
= Z0 �ZR0� ⇒ Zin = R
3
5. A conductor having a cross-sectional area a sq. m carrying
current I ⃗J A, ties in a magnetic field
B�⃗ = B0(⃗ı + ⃗ȷ)Wb/m2
The force density on the conductor is
where ⃗ , ⃗ and � ⃗ are orthogonal unit vectors.
a) B0 Ik�⃗
a
b) − B0 I⃗ı
a
c) − B0 I�k⃗
a
d) B0 I⃗ȷ
a
Answer: (c)
Solution:
⃗I = I⃗ȷA
B�⃗ = B0(ı̂ + ȷ)̂ Wb/m2
F�⃗ = l�⃗I × B�⃗�
= l{I⃗ȷ × B0(i + ȷ)̂ }
Force density, F = − B0 Ik�
vol a
6. A generator of 50 Ω internal impedance and operating at 1 GHz
feeds a 75 Ω load via a coaxial line of characteristic impedance
50 Ω. The voltage standing wave ratio on the feed line is
a) 0.50
b) 1.50
c) 1.75
d) 2.50
4
Answer: (b)
Solution:
VSWR = 1 + |ΓL|
1 − |ΓL|
ΓL = ZL − Z0
ZL + Z0
ZL = 25Ω, Z0 = 50Ω
ΓL = 75 − 50 = 1
75 + 50 5
VSWR = 1+15 = 3 = 1.5
1−15 2
7. If the current density inside a straight conductor is uniform over
its cross-section, the flux density variation inside the conductor
at different distances from its centre is
a) linear
b) square of the distance
c) inverse of the distance
d) exponential
Answer: (a)
Solution:
For 0 < r < R,
∮ . = ∫ .
5
= .
2
= . .
2
8. Given that: ∇ × H��⃗ = ⃗J + ∂�D�⃗
∂t
Assertion (A): In the equation, the additional term ∂D��⃗ is
∂t
necessary.
Reason (R): The equation will be consistent with the principle
of conservation of charge.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is NOT the correct explanation
of A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (a)
Solution:
∂D��⃗ refers to displacement current through the dielectric medium.
∂t
Presence of this in the Faraday’s law conforms to the principle
of conservation of charge.
9. Divergence of the three-dimensional radial vector field F�⃗ is
a) 3
b) 1/r
c) ı̂ + ȷ̂ + k�
d) 3(ı̂ + ȷ̂ + k� )
Answer: (a)
6
Solution:
r⃗ = xa���x⃗ + ya����y⃗ + z�a��z⃗
div(r⃗) = (∇. r⃗) = ∂ x + ∂ y + ∂ z
∂x ∂y ∂z
=1+1+1=3
10. Match List-I (Expression) with List-II (Coefficient) and
select the correct answer using the code given below:
List-I
A. ρ = ZL−Z0
ZL+Z0
B. ρ = Z0−ZL
ZL+Z0
C. ρ = 2ZL
Z0+ZL
D. ρ = 2Z0
Z0+ZL
List-II
1. Reflection coefficient for voltage
2. Transmission coefficient for voltage
3. Reflection coefficient for current
4. Transmission coefficient for current
5. Standing-wave ratio
Codes:
ABC D
a) 1 2 5 4
b) 4 3 2 1
c) 1 3 2 4
d) 4 2 5 1
7
Answer: (c)
Solution:
Reflection coefficient of voltage = ρV = ZL−Z0
ZL+Z0
Current reflection coefficient = negative of voltage
Reflection coefficient = Z0−ZL
ZL+Z0
Transmission coefficient for voltage = 1 + ρV = 2ZL
ZL+Z0
Transmission coefficient for voltage =ZL2+Z0Z0
A–1
B–3
C–2
D–4
11. Maxwell equation �∇⃗ × �E⃗ = − �∂∂�B�t⃗� is represented in integral
form as
a) ∮ �E⃗. �d��⃗ = − ∂ ∮ �B⃗. �d��⃗
∂t
b) ∮ �E⃗. �d��⃗ = − ∂ ∫S �B⃗. �d���s⃗
∂t
c) ∮ �E⃗ × �d��⃗ = − ∂ ∮ B�⃗. �d��⃗
∂t
d) ∮ �E⃗ × �d��⃗ = − ∂ ∫S B�⃗. �d��⃗
∂t
Answer: (b)
Solution:
∇�⃗ × �E⃗ = − �∂∂B��t⃗�
Taking integration on both sides on a surface � � � � ⃗
8
∫S ∇. E�⃗. d����s⃗ = − ∂ ∫S �B⃗. d����s⃗ …….. (1)
∂t
But according to stokes theorem
∮ �E⃗. �d��⃗l = ∫S curl�E⃗. �d���s⃗ = ∫S ∇ × �E⃗. d����s⃗ ……… (2)
So from equation (1) and (2) we have
∮ �E⃗. �d��⃗ = − ∂ ∫S �B⃗. �d���s⃗
∂t
12. In free space, if ρ = 0, the Poisson’s equation becomes
a) Maxwell s divergence equation V.E = 0
b) Laplacian equation ∇2V = 0
c) Kirchhoff s voltage equation ΣV = 0
d) None of the above
Answer: (b)
Solution:
For a homogeneous medium Poisson’s equation is
∇2V = − ρv
ϵ
If ρv = 0 (charge free region)
∇2V = 0 this is laplace equation
13. For a parallel plate transmission line with perfectly conducting
plates of width w and separated by a lossless dielectric slab of
thickness d, the characteristic impedance Z0 is p times the
intrinsic impedance η of the dielectric medium where
a) p = d
w
b) p = w
d
9
c) p = �wd
d) p = �wd
Answer: (a)
Solution:
For parallel plate line,
0 = � . Thickness
Width
= � = η
Also, 0 = η (Given)
So, P = d
w
C = ϵd F/m
w
L = μw
d
Z0 = �CL = �μϵ . �wd22
= �μϵ d
w
10
14. Consider the following three equations:
1. ∇ × �E⃗ = − ∂B��⃗
∂t
2. ∇ × �H�⃗ = ⃗J + ∂D��⃗ 0
∂t
3. ∇. �B⃗ = 0
Which of the above appear in Maxwell’s equations?
a) 1, 2 and 3
b) 1 and 2
c) 2 and 3
d) 1 and 3
Answer: (a)
Solution:
∇ × �E⃗ = − ∂B��⃗
∂t
Maxwell’s 3rd equation is derived from Faraday’s law of
electromagnetic induction.
∇ × �H�⃗ = ⃗J + ∂D��⃗
∂t
Maxwell’s 4th equation is generalized form of Ampere’s law.
∇. �B⃗ = 0
Maxwell’s equation is also Gauss law for Magnetic fields.
15. An isotropic antenna is radiating 3 kW. The field at 3 km
distance is
a) 0.1 V/m
b) 10 V/m
c) 0.01 V/m
11
d) 0.2 V/m
Answer: (a)
Solution:
Power flow per unit area = P = E2
4πr2 η0
E → Field, η0 = 120π
P = 3 × 103 W
r = 3 × 103 m
3×103 = E2
(4π)(3×103)2 120π
E2 = 0.01
E = 0.1 V/m
16. The inconsistency of continuity equation for time varying
fields was corrected by Maxwell and the correction applied was
a) Ampere’s law, ∂�D�⃗
∂t
b) Gauss’s law, ⃗J
c) Faraday’s law, ∂�B�⃗
∂t
d) Ampere’s law, ∂�P⃗
∂t
Answer: (a)
Solution:
Modified ampere’s law:
∇ × �H�⃗ = ⃗J + ∂D��⃗
∂t
12
17. A flat slab of dielectric, εr = 5 is placed normal to a uniform
field with a flux density D = 1 coulomb/m2. The slab is
uniformly polarized. What is the polarization P of the slab in
coulomb/m2?
a) 0.8
b) 1.2
c) 4
d) 6
Answer: (a)
18. A current carrying conductor of length l is under the influence
of a magnetic field having magnetic flux density B. If I is the
current flowing through the conductor, which of the following
formula is correct for calculating the force exerted on it?
a) F = B × × I2
b) F = B×I
c) F = I×
B
d) F = B × I ×
Answer: (d)
19. A parallel plate waveguide has a plate spacing of 10 mm and is
filled completely with a dielectric medium of relative
permittivity 9.0. The cut-off frequency of the first (lowest) TE
mode is
a) 1.0 GHz
b) 2.5 GHz
13
c) 5.0 GHz
d) 10.0 GHz
Answer: (c)
Solution:
fc = 1 ��ma �2 + �nb�2
2√μϵ
Lowest order TE mode is TE10
m = 1, n = 0, a = 10 × 10-3, ϵr = 9, μ = μ0, ϵ = ϵ0ϵr
fc = 1 × 1 ��10×110−3�2
2�μ0ϵ0 √9
= 3×108
(2)(3)(10×10−3)
= 5GHz
20. Which one of the following formulae is NOT correct for the
boundary between two magnetic materials?
a) 1 = 2
b) 2 = � 2 + 2
c) 1 = 1 + 1
d) � 21 × � � ⃗ 1 − � ⃗ 2� =
where � 21 is a unit vector normal to the interface and directed
from region 2 to region 1.
Answer: (b)
14
21. In a rectangular waveguide, what is the ratio of wave
impedance for TM mode to wave impedance for TE mode
(i.e. XTM⁄ZTE)?
a) 1
b) 1
[1−(fc⁄f)2]
c) [1 − (fc⁄f)2]
d) [1 − (fc⁄f) ]
Answer: (c)
Solution:
= � �1 − � �2
= �
�1−� �2
= �1 − � �2�
22. Consider a solid sphere, of radius 5 cm made of a perfect
electric conductor. If one million electrons are added to this
sphere, these electrons will be distributed
a) uniformly over the entire volume of the sphere
b) uniformly over the outer surface of the sphere
c) concentrated around the centre of the sphere
d) along a straight line passing through the centre of the sphere
Answer: (b)
15
Solution:
Even if we give any number of charges, the charge will reside
on its surface only.
23. Which one of the following statements is correct? For a loss
less dielectric medium, the phase constant for a travelling wave,
β is proportional to
a) ϵr
b) √ϵr
c) 1
ϵr
d) 1
√ϵr
Answer: (b)
24. For similar field distribution (not identical) the highest Q
factor can be obtained from which one of the following cavities?
a) Rectangular cavity
b) Circular cavity
c) Spherical cavity
d) Co-axial cavity
Answer: (c)
Solution:
Q = 2 . volume
8 surface area
16
For all shapes Volume/Surface area is highest for sphere. So
spherical cavity has highest Q - factor.
25. Magnetic flux density at a point distance R due to an infinitely
long linear conductor carrying a current I is given by
a) B = I
2πR
b) B = μI
2R
c) B = μI
2πR
d) B = μI
2πR2
Answer: (c)
26. In a waveguide, the guide wavelength (λg) for TEmn and
TMmn, modes at frequency above the cut-off follows which
relation?
a)(λg) TEmn < TMmn
b)(λg) TEmn > (λg) TMmn
c)(λg) TEmn = (λg) TMmn
d) The relation depends on specific values of m and n
Answer: (c)
Solution:
λg = 2
��ma �2+�bn�2
For TE lowest mode is TE10 or TE01
For TM lowest mode is TM11
�λg�TE10 = 2 = 2a
�a12
17
�λg�TE11 = 2
�a12+b12
�λg�(TE10 or TE01) > �λg�TM11
But for a specific m and n,
�λg�TEmn = �λg�TMmn
27. For an ideal multi-conductor transmission line with
propagation along the z-axis, the electric and magnetic fields
between and on the conductors of the line are characterized by
a) Ex = 0, Ey = 0, Ex ≠ 0
b) Ex = 0, Hz = 0
c) Ex ≠ 0, Ey ≠ 0, Ez = 0
d) Ex ≠ 0, Hy ≠ 0, Ez > 0
Answer: (b)
28. If the radiation electric field at the point (r, θ, ϕ) due to a short
dipole antenna located at the origin is E then what would be the
field at the point (2r, θ, ϕ)?
a) E/8
b) E/4
c) E/2
d) E/1
Answer: (c)
Solution:
For far field or radiation field ∝ 1
Field at r → E Field at 2r → E
2
18
29. A parallel plate capacitor with air as dielectric can withstand a
voltage V. If the capacitor is half filled with a liquid of dielectric
constant 2, then the capacitor will withstand
a) 2 V
b) 4 V
3
c) 3 V
4
d) 1 V
2
Answer: (c)
30. The flux density at a point in space is given by B�⃗ = 4xa�x +
2kya�y + 8a�zWb/m2. The value of constant k must be equal to
a) –2
b) –0.5
c) +0.5
d) +2
Answer: (a)
Solution:
Magnetic flux density is solenoidal field i.e., for free space
∇. � = 0
⇒ ∂ (4x) + ∂ (2ky) + ∂ (8) = 0
∂x ∂y ∂z
⇒ 4 + 2k = 0 ⇒ k = −2
19
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