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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