Electromagnetic Fields Test - 5 - PDF Flipbook

Electromagnetic Fields Test - 5

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GATE
EEE

Electro
MagneticFields

Test-05Solutions


ELECTROMAGNETIC FIELDS

1. If a vector field B is solenoidal, which of these is true?
a) ∮L B�⃗ ∙ �d��⃗ = 0
b) ∮S �B⃗ ∙ �d���s⃗ = 0
c) ∇ × �B⃗ = 0
d) ∇ ∙ �B⃗ ≠ 0
Answer: (b)
Solution:
According to gauss law for magnetic field
∇ ∙ �B⃗ = 0 i.e., no existence of monopoles
∮S �B⃗ ∙ �d���s⃗ = 0 (∵ Divergence theorem)
(∵ divergence of �B⃗ = 0 i.e., B�⃗ is solenoidal

2. Match List-I (Quantity) with List-II (Expression) and select
the correct answer:
List – I
A. Power flow density
B. Impedance of media
C. Joule heating
D. Velocity of a light
List – II
1. �E⃗ × �H�⃗
2. �E⃗/�H�⃗
3. �E⃗. H��⃗

1


4. �E⃗/�B⃗

Codes:

ABC D

a) 1 2 3 4

b) 4 3 2 1

c) 1 3 2 4

d) 4 3 2 1

Answer: (a)

Solution:

By Poynting theorem,
PD = �E⃗ × �H�⃗ (Power flow density)

Characteristic impedance,

η = E� = �μϵ
H�

Also, B� = μH�

E� = E� = 1 �μϵ = 1 (velocity of light)
B� μH� μ √μϵ

3. Which one of the following is the correct statement?

Equipotential lines and field lines

a) are parallel

b) are anti-parallel

c) are orthogonal

d) bear no definite relationship

Answer: (c)

2


Solution:
Equipotential lines: Suppose scalar potential V is function of Z
so for a particular Z, a plane parallel to xy plane will have
equipotential lines, we know that E�⃗ = −∇V, gradient of scalar
potential V will give a line perpendicular to plane on which V
lie i.e., perpendicular to xy plane so equipotential lines and field
lines are orthogonal to each other.
4. If E = 0 at all points on a closed surface,
1. The electric flux through the surface is zero.
2. The total charge enclosed by the surface is zero.
3. Charge resides on the surface.
a) 1 and 2 only
b) 1 and 3 only
c) 2 and 3 only
d) 1, 2 and 3
Answer: (a)
5. An electric field in a charged medium with a time varying
magnetic field has
a) ∇�. E� = 0; ∇� × E� = 0
b) ∇�. E� ≠ 0; ∇� × E� ≠ 0
c) ∇�. E� = 0; ∇� × E� ≠ 0
d) ∇�. E� ≠ 0; ∇� × E� = 0
Answer: (b)

3


Solution

In a charged medium and time varying magnetic field,

∇. E� = ρv
ϵ

So, ∇. E� ≠ 0

∇ × E� = −μ ∂H� ;
∂t

∂H� ≠ 0
∂t

So, ∇ × E� ≠ 0

6. According to Gauss's Law, the surface integral of the normal

component of electric flux density D over a closed surface

containing charge Q is

a) Q
ε0

b) ε0Q

c) Q

d) Q2
ε0

Answer: (c)

Solution:
Gauss’s law states that the total electric flux Ψ through any

closed surface is equal to the total charge enclosed by that

surface

Ψ =
= ∮ .

4


7. Which of the following represent the Maxwell’s curl equation

for static magnetic field?

a) ∇ × �B⃗ = μ0⃗ȷ
b) ∇ × �B⃗ = 0
c) ∇. �B⃗ = μ0
d) ∇. �B⃗ = 0

Answer: (a)

Solution:

As per ampere's circuital law,

∮ B. dl = μ0 ∬ J. dS
= ∬(∇ × B)dS = μ0 ∬ J. dS
= ∇ × B = μ0J

8. The capacitance of a concentric spherical capacitor of she radii x

and y (x × y) is

a) 1 ln x
4πε0 y

b) 4πε0xy
x−y

c) 4πε0ln y
x

d) 1 �y1 − 1x�
4πε0

Answer: (b)

Solution:

V = Q �y1 − 1x�
4πcos

Q = CV

5


C = Q = Q
V 4πQcos�y1−1x�

= 4πcosxy
x−y

9. Assertion (A): The relationship between Magnetic Vector
potential ⃗ and the current density J in free space is ∇ ×
�∇ × �A⃗� = μ0⃗J .

For a magnetic field in free space due to a dc or slowly varying
current is ∇2A�⃗ = −μ0⃗J
Reason (R): For magnetic field due to dc or slowly varying
current ∇ ∙ ⃗ = 0.

a) Both A and B 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:
∇ × ∇ × �A⃗ = ∇�∇. A�⃗� − ∇2�A⃗; for steady

magnetic field ∇. �A⃗ = 0; ∇2A�⃗ = −μ0⃗J
(Poisson’s equation) so that ∇ × ∇ × �A⃗ = μ0⃗J

6


10. The magnetic vector potential at the point P due to the current
loop C shown in the below figure is in the direction of

a) �a⃗x

b) �a⃗y

c) �a⃗z

d) �a⃗x+a�⃗y
√2

Answer: (b)

Solution:

Magnetic vector potential due to line charge

A� = μ0I ∮ dl
4π R

d�l = (a. dϕ)a�0 (Cylindrical coordinates)
Also conservation of a�ϕ into Cartesian coordinates

a�ϕ = −a�x sin ϕ + a�y cos ϕ

7


But for point P, ϕ = 0

a�ϕ = a�y
So, direction of A� at P is a�y.

11. In a source-free imperfect dielectric medium (specified by loss

tangent tan δ), Maxwell's curl equation can be written as:
a) ∇ × H� = jωεE�(1 + jtanδ)
b) ∇ × H� = jωεE�(1 − jtanδ)
c) ∇ × H� = −jωεE�(1 + jtanδ)
d) ∇ × H� = −jωεE�(1 − jtanδ)

Answer: (b)

Solution:

For a lossy dielectric Maxwell an equation

∇ × H� = (σ + jωε)E�
Also, tanδ = ωσε,
So, ∇ × H� = jωε �1 − j ωσε� E�

= jωε(1 − jtanδ)E�

12. A solid cylindrical conductor of radius ‘R’ carrying a current 'I'

has a uniform current density. The magnetic field intensity ‘H’

inside the conductor at the radial distance (r < R) is

a) Zero

b) I
2πr

c) Ir
2πR2

d) IR2
2πR3

8


Answer: (c)

Solution:

In general �H�⃗ = I a� ϕ , Current enclosed at distance
2πr

r = I × πr2 = Ir2
πR2 R2

∴ �H�⃗(r) = Ir2 = Ir a� ϕ
2πr.R2 2πR2

13. A plane wave propagates in a direction having direction

cosines (1/√2, 1/√2, 0). The equation of the phase fronts is

a) xy = constant

b) x/y = constant

c) x + y = constant

d) x – y = constant

Answer: (c)

Solution:

Phase front is a plane perpendicular to the direction of wave

motion.

Let n� be a unit vector in the direction of wave propagation.

Here, n� = 1 a� x + 1 a� y
√2 √2

P is any point on phase front

9


O���P� = r̅ = xa�x + ya�y + za�z
Equation of phase front is n�. r̅ = constant

�√12 a� x + 1 a� y � �xa� x + ya� y + za� z �= constant
√2

x + y = constant
√2 √2

Or x + y = constant

14. Two conducting coils 1 and 2 (identical except that 2 is split)

are placed in a uniform magnetic field which decreases at a

constant rate as in the figure. If the planes of the coils are

perpendicular to the field lines, the following statements are

made:

1. an e.m.f is induced in the split coil 2
2. e.m.f s are induced in both coils
3. equal Joule heating occurs in both coils
4. Joule heating does not occur in any coil
Which of the above statements is/are true?
a) 1 and 4
b) 2 and 4
c) 3 only
d) 2 only
Answer: (d)

10


Solution:
Emf will be induced in both the coil due to rate of change of
magnetic field. Since coil 2 is split so no current will flow in it
so no Joule heating but in coil 1 Joule heating will occurs as
current flows through it.
15. Match the relationship of items given in List-I (Medium) with
the List-II (Expression for intrinsic impedance) and select the
correct answer using the code given below:
All symbols have their usual meanings:
List-I
A. Free space
B. Perfect dielectric
C. Partially conducting medium
D. Conducting medium
List – II
1. �ωσμ . ∠450
2. �σj+ωjωμ ε
3. �με
4. �με00

11


Codes:
ABC D

a) 1 2 3 4
b) 1 3 4 2
c) 4 2 1 3
d) 4 3 2 1
Answer: (d)
Solution:
Free space = �με00

Perfect dielectric = �με

Partially conducting medium or lossy dielectric = �σj+ωjωμε

Conducting medium = �ωσμ ∠450

16. What is the force experienced per unit length by a conductor
carrying 5 A current in positive Z-direction and placed in a
magnetic field B = (3�a⃗x + 4a�⃗y)?
a) 15�a⃗x + 20�a⃗y N/m
b) −20�a⃗x + 15�a⃗y N/m
c) 20�a⃗x − 15�a⃗y N/m
d) −20�a⃗x − 20�a⃗y N/m
Answer: (b)
Solution:
Force on current carrying conductor due to magnetic field

12


�B⃗ = (3a�x + 4a�y)
F�⃗ = I(a�z + �B⃗)

Force per unit length F�⃗1 = I(a�z + �B⃗)
= 5�a�z × �3a�x + 4a�y��

= 5�3a�y − 4a�x�
F�⃗1 = −20a�x + 15a�y N/m

17. Assertion (A): A low voltage standing wave ratio (VSWR) is

the goal in a transmission line.

Reason (R): The higher the VSWR, the greater is the mismatch

on the line.

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 a correct explanation of

A.

c) A is true, but R is false.

d) A is false, but R is true.

Answer: (a)

Solution:

Voltage standing wave ratio VSWR = Vmax
Vmin

Smin = 1. So a low VSWR implies value of s near to 1.Smax = ∞

A matched line implies that, Zin = Z0(ZL = Z0)

Zin = Z0 for ZL = Z0

Zin = Z0 �ZZL0++jjZZL0 tan ββ �
tan

Zin = Z0

13


ΓL(Reflection coeff. ) = ZL−Z0 = 0
ZL+Z0

s = 1+|ΓL|
1−|ΓL|

So, line is matched for s = 1 with increase in s, mismatch

increases.

• ZL = 0 gives Zin = jZ0 tan β and s = ∞
• ZL = ∞ gives Zin = −jZ0 cot β and s = ∞
18. Which of the following is zero as applied to electromagnetic

fields?
a) grad div �A⃗

b) div grad V
c) div curl �A⃗
d) curl curl A�⃗

Answer: (c)

Solution:
∇. �∇ × �A⃗� i.e., Div curl A�⃗ is always zero

and ∇ × (∇ϕ) i.e., curl gradϕ is always zero

19. Which of the following equations is/are not Maxwell's

equation(s)?

1. ∇ ∙ ⃗J = − ∂ρv
∂t

2. ∇ ∙ �D�⃗ = ρv

3. ∇. �E⃗ = − ∂�B�⃗
∂t

4. ∮ �H�⃗. d���⃗l = ∫S �σ�E⃗ + ε ∂∂E�t⃗� . d����S⃗

14


Select the correct answer using the codes given below:
a) 2 and 4
b) 1 alone
c) 1 and 3
d) 1 and 4
Answer: (c)

Solution:

∇ ∙ ⃗J = − ∂ρv is not a Maxwell’s equation
∂t

as ∇ ∙ �D�⃗ = ρv is a Maxwell’s first equation

∇ × �E⃗ = − ∂B��⃗ is Maxwell’s third equation
∂t

So, ∇. E�⃗ = − ∂B��⃗ is not a Maxwell’s equation
∂t

20. If the cut-off frequency of a parallel plate wave guide for TE10

mode is 2000 MHz, the cut-off frequency for TE20 mode will be

a) 1000 MHz

b) 2000 MHz

c) 3000 MHz

d) 4000 MHz

Answer: (d)

Solution

Cut-off frequency, fc = m . v
2a

a → separation between plate

v → velocity

15


fc1 ∝ m1
fc2 ∝ m2
m → mode

For TE10 m = 1

TE20 m = 2

fc2 = m2 = 2
fc1 m1 1

⇒ fc2 = 2fc1 = (2) (2000) = 4000 MHz

21. Which one of the following equations is not Maxwell's

equation for a static electro-magnetic field in a linear

homogeneous medium?
a) ∇. �B⃗ = 0
b) ∇ × �D�⃗ = 0
c) ∮C �B⃗. �d��⃗l = μ0I
d) ∇2�A⃗ = μ0⃗J
Answer: (b)

Solution:

∇. �B⃗ = 0 Maxwell’s 2nd equation

∇ × �D�⃗ = 0 is not a Maxwell’s equation

∇ × �D�⃗ = ρv is Maxwell’s 1st equation

∇ × �E⃗ = − ∂B��⃗ is Maxwell’s 3rd equation
∂t

∇ × �H�⃗ = ⃗JC + ∂D��⃗ is Maxwell’s 4th equation
∂t

16


22. Assertion (A): A wave guide does not support TEM wave.
Reason (R): For TEM wave to exist on a transmission line, an
axial conduction current must be supported.
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 a correct explanation of
A.
c) A is true, but R is false.
d) A is false, but R is true.
Answer: (d)
Solution:
A waveguide may or may not support TEM depending upon
design like plate separation, dielectric constant of material
inside. Transmission line support TEM due to conduction.

23. Silicon steel is used in electrical machines because it has
a) low hysteresis loss
b) low retentivity
c) low coercivity
d) none of these
Answer: (a)

24. The magnitude of emf induced in a wire does not depend on
a) length of wire
b) speed of wire
c) diameter of wire
d) none of these
Answer: (c)

17


25. Which one of the following gives the approximate value of

capacitance between two spheres, whose separation is very

much larger than their radii R?

a) 2π
ϵ0R

b) 2πϵ0R

c) 2πϵ0
R

d) 4πϵ0
R

Answer: (b)

26. In a rectangular waveguide of cross section a×b, the dimension

'a' determines the cut-off wavelength. What does the dimension

'b' determine?

a) The bandwidth

b) Power handling capacity

c) The propagating mode

d) The dispersion characteristic

Answer: (b)

Solution:

λc = 2
��ma �2+�bn�2

as λc depends upon ‘a’ only, so n = 0

It is a TE10 mode as TM10 not possible.

Power Pavg for TE10 = ω2μ2a3μ02b
4π2η

So, Pavg ∝ b

18


27. When the load impedance is equal to the characteristic

impedance of the transmission lines then the reflection

coefficient and standing wave ratio are respectively

a) 0 and 0

b) 1 and 0

c) 0 and 1

d) 1 and 1

Answer: (c)

28. A capacitor is made up of two concentric spherical shell. The

radii of the inner and outer shells are R1 and R2 respectively and

ε is the permittivity of the medium between the shells. The

capacitance of the capacitor is given by

a) 1
4πε�R11−R12�

b) 1 �R11 − R12�
4πε

c) 4πε R1R2
R1+R2

d) 4πε R1R2
R1−R2

Answer: (d)

29. The far-field intensities of a magnetic dipole antenna vary

inversely as

a) R3

b) R2

c) R

d) √R

19


Where R is the radial distance of the observation point from the

center of the loop.

Answer: (c)

Solution:

For far field or radiation field

∝ 1


30. Which one of the following statements is correct?

A wave guide can be considered to be analogous to a

a) low pass filter

b) high pass filter

c) band pass filter

d) band stop filter

Answer: (b)

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