Electromagnetic Fields Test - 4 - PDF Flipbook

Electromagnetic Fields Test - 4

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GATE
EEE

Electro
MagneticFields

Test-04Solutions


ELECTROMAGNETIC FIELDS
1. An electromagnetic field is radiated from

a) a stationary point charge
b) a capacitor with a DC voltage
c) a conductor carrying a DC current
d) an oscillating dipole
Answer: (d)
Solution:
A dipole is an antenna. A dipole to which an alternating current
is fed is known as an oscillating dipole, which radiates
electromagnetic field. The other options refer to stationary
charges, or steady currents, which do not produce radiation.
2. Image theory is applicable to problems involving
a) electrostatic field only
b) magnetostatic field only
c) both electrostatic and magnetostatic fields
d) neither electrostatic nor magnetostatic field
Answer: (a)
Solution:
Image theory is applicable to problem involving electrostatic
fields only normally to find the field on a conductor surface
(Normal component).

1


Statement for Linked Answer Q.3:

An inductor designed with 400 turns coil wound on an iron core

of 16 cm2 cross sectional area and with a cut of an air gap length

of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply.

Neglect coil resistance, core loss, iron reluctance and leakage

inductance (μ0 = 4π × 10-7 H/m).

3. The current in the inductor is

a) 18.08 A

b) 9.04 A

c) 4.56 A

d) 2.28 A

Answer: (d)

Solution:

Reluctance of the iron part of the flux is neglected. Hence

reluctance of the flux path.

= 10−3 = 10−3 = 108 AT/Wb
μ016×10−4 4π10−7×16×10−4 64π

For a current I A in coil; mmf = 400 I AT;

Flux = 400 I×64π Wb
108

Inductance of coil = 4002×64π H = 1024π = 0.3217H
108 104

(No leakage, no fringing, no resistance, no core loss), X = ωL

Reactance = 100 π (0.3217) = 101.06 Ω

Coil current, i = 230/101.06 = 2.276 A

2


4. Which one of the following pairs is NOT correctly matched?

a) Gauss Theorem: ∮S �D�⃗ ∙ d����s⃗ = ∮V ∇ ∙ �D�⃗ dv

b) Gauss Theorem: ∮ �D�⃗ ∙ �d���s⃗ = ∮V ρ dv

c) Coulombs Law: V = − d∅m
dt

d) Stoke’s Theorem: ∮l �E⃗ ∙ d���⃗l = ∮S(∇ × �E⃗) ∙ �d���s⃗

Answer: (c)

Solution:
It is Faraday’s Law �V = − dd∅tm�
5. Which one of the following statements is correct?

The polarizability of a conducing metallic sphere is

a) proportional to the cube of the radius of the sphere

b) proportional to the radius of the sphere.

c) cannot be determined as the sphere is metallic.

d) independent of the dimensions of the metallic sphere

Answer: (a)

Solution:
αe = 4πϵ0R3

αe → Electric polarizability

R → Radius of sphere
αe ∝ R3

6. Ohm's law in point form in field theory can be expressed as

a) V = RI

b) J = E/σ

3


c) J = σE

d) R = ρl/A

Answer: (c)

Solution:
⃗ = σ � ⃗ is point form of ohm’s law

7. A circular disc of radius R carries a uniform surface charge

density. When it revolves at a uniform angular velocity about its

Centre and in its own plane, the magnetic flux density at the

Centre of the disc is B. If the radius of the disc is doubled and

the original charge spread out uniformly on the extended area,

the magnetic field at the Centre would be

a) B/4

b) B/2

c) B

d) 2B

Answer: (b)

Solution:

Magnetic field due to charge disc with surface charge density

‘ 1’, radius ‘R1’

B1 = μ0σ1ωR1

σ1 = Q (Q → charge)
πR21

For, R2 = 2R1

σ2 = Q = Q = σ1
πR22 π(2R1)2 4

4


B2 = μ0σ2ωR2

= μ0 �σ41� (ω)(2R1)

= 1 μ0σ1ωR1 = 1 B1
2 2

8. In cylindrical coordinate system, the potential produced by a

uniform ring charge is given by ∅ = f(r, z), where f is a

continuous function of r and z. Let � ⃗ be the resulting electric

field. Then the magnitude of ∇ × E�⃗

a) increase with r

b) is 0

c) is 3

d) decrease with z

Answer: (b)

Solution:

A uniformly charged ring is specified. It can be considered as

static. A static electric charge produces an electric field for
which ∇ × � = 0

9. A square loop and an infinitely long conductor, each carries a

current I as shown in the figure given below. What is the force

on the loop?

5


a) μ0I2 Away from the conductor.


b) μ0I2 Towards to the conductor.


c) μ0I2 loge 2 Away from the conductor.


d) μ0I2 loge 2 Towards to the conductor.


Answer: (a)

Solution:

Force, F� = I(l × B�)

dF�1 = I(dl × B�)

B� at x → μ0 I a� y ,
4π x

d� = dxa�x

dF�1 = I �4μπ0 I dx� a� z
x

dF�1 = ∫ dF�1 = μ0I2 a� z ∫a2a dx
4π x

F�1 = μ0I2 ln2 a� z


Similarly, F�3 = − μ0I2 ln2 a� z


F�1 + F�3 = 0

6


F�4 = (I)(a) μ0 . I . a�x = μ0I2 a� x
2π a 2π

F�2 = −(I)(a) �2μπ0 . 2Ia� . a�x

= − μ0 I2 a� x


F� = F�2 + F�4

= μ0I2 . a� x (Away from conductor)


10. Point charges of Q1 = 2 nC and Q2 = 3 nC are located at a

distance apart. With regard to this situation, which one of the

following statements is not correct?

a) The force on the 3 nC charge is repulsive.

b) A charge of -5 nC placed midway between Q1 and Q2 will

experience no force.

c) The forces Q1 and Q2 are same in magnitude.

d) The forces on Q1 and Q2 will depend on the medium in which

they are placed

Answer: (b)

Solution:

Q1 = 2 nC, Q2 = 3 nC, Q3 = -5 nC

Force on Q3 by Q1

F13 = 1 Q1Q3
4πϵ0 r2

= 1 −10×10−9×10−9
4πϵ0 d2/4

= 1 −40×10−9×10−9
4πϵ0 d2

7


Force on Q3 and Q2

F23 = 1 Q2Q3
4πϵ0 r2

= 1 −15×10−9×10−9
4πϵ0 d2/4

= 1 −60×10−9×10−9
4πϵ0 d2

Total force F = �F⃗13 + F�⃗23 ≠ 0

So option (b) is wrong

11. A quantitative relation between induced emf and rate of

change of flux linkage is known as

a) Maxwell's law

b) Stoke’s law

c) Lenz’s law

d) Faraday’s law

Answer: (d)

Solution:

Faraday's law of electromagnetic induction says that magnitude

of induced emf is directly proportional to the rate of change of

flux.

12. If the magnitude of the magnetic flux B at a distance of 1 m

from an infinitely long straight filamentary conducting wire is 2

× 10-6 Wb/m2, what is the current in the wire?

a) 1 A

b) 10 A

c) 100 A

d) 1000 A

8


Answer: (b)

Solution:

Magnetic field density due to infinite wire,

B0 = μ0I
2πa

Given, a = 1 m

B = 2 × 10-6 Wb/m2

μ0 = 4π × 10-7

2 × 10-6 = (4π)�10−7� . I = 2 × 10−7J
2π 1

⇒ I = 10 Amp

13. The electric field lines and equipotential lines

a) are parallel to each other

b) are one and the same

c) cut each other orthogonally

d) can be inclined to each other at any angle

Answer: (c)

Solution:

The electric field lines and equipotential lines cut each other

orthogonally.

9


14. For the following statements associated with the basic
electrostatic properties of ideal conductors:
1. The resultant field inside is zero.
2. The net charge density in the interior is zero.
3. Any net charges reside on the surface.
4. The surface is always equipotential.
5. The field just outside is zero.
Which of the above statements are correct?
a) 1, 2, 3 and 4
b) 3, 4 and 5 only
c) 1, 2 and 3 only
d) 2 and 3 only
Answer: (a)
Solution:
The basic electrostatic properties of idea conductors are:
1. the resultant field inside is zero.
2. the net charge density in the interior is zero.
3. any net charges reside on the surface.
4. the surface is always equipotential.
The field just outside the conductor is not zero.

15. A hollow metallic sphere of radius r is kept at potential of 1
Volt. The total electric flux coming out of the concentric
spherical surface of radius R (> r) is
a) 4πε0r
b) 4πε0r2

10


c) 4πε0R
d) 4πε0R2
Answer: (a)
16. For the current I decreasing in the indicated direction, the
e.m.f. in the two loops A and B shown in the figure below, is in
the direction

a) clockwise in A and anticlockwise in B
b) anticlockwise in A and clockwise in B
c) clockwise in both A and B
d) anticlockwise in both A and B
Answer: (b)
Solution
Direction of induced emf and consequent current should be such
that field produced by this current in loops strengthen the field
produced by ‘I’ (Lenz law). Anticlockwise in 'A' and clockwise
in 'B' produces the field that strengthen the field by 'I'.
17. The field strength at a point of finite distance from an infinitely
long straight uniformly charged conductor is obtained by
considering the radial (R) component and the longitudinal (L)
component of the forces acting on a unit charge at the point, by
the charges on the elemental length of the conductor. The
resultant field strength is

11


a) the sum of R-components, when the sum of L-components is
zero

b) the sum of L-components, when the sum of R-components is
zero

c) the sum of both R- and L-components
d) average of the sums of R- and L-components
Answer: (a)
Solution:

According to coulomb’s law the force or field strength at point

‘P’ will act only in radial (R) direction while longitudinal

components (L) direction while longitudinal components (L)

will get cancelled out.

18. Plane defined by z = 0 carry surface current density 2a�x A/m.
The magnetic intensity ‘Hy’ in the two regions – α < z < 0 and 0

< z < α are respectively

a) a�y and − a�y
b) −a�y and a�y
c) a�x and − a�x
d) −a�x and a�x

12


Answer: (a)

Solution:

for – α < z < 0 H� y = 1 (2)a�y = a� y
2

for 0 < z < α H� y = 1 (2)�−a�y� = a� y
2

19. What does a time-rate of change of electric displacement lead

to?

a) Convection current

b) Conduction current

c) Displacement current

d) No current flow

Answer: (c)

Solution:

Maxwell’s equation for time varying field,

∇ × H� = J̅ + ∂D�
∂t

J̅ → Conduction current density

∂D = Jd̅ → Displacement current density
∂t

20. Assertion (A): ∫S �B⃗ ∙ �d���s⃗ = 0 where, � ⃗ is magnetic flux density,

d����s⃗ = vector with direction normal to surface element ds.

Reason (R): Tubes of magnetic flux have no sources or sinks.

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

13


d) A is false but R is true

Answer: (a)

Solution:
∫S �B⃗ ∙ �d���s⃗ = 0

Maxwell’s 2nd equation → i.e., single pole does not exist

because there is no source or sink separately so both are correct

and R is correct explanation for A.

21. The direction of the magnetic lines of forces is

a) from + to – charges

b) from south to north poles

c) from one end of the magnet to the other

d) from north to south poles

Answer: (d)

22. Which one of the following is not the valid expression for
magnetostatic field vector � ⃗ ?

a) �B⃗ = ∇ ∙ �A⃗
b) �B⃗ = ∇ × �A⃗
c) ∇ ∙ �B⃗ = 0
d) ∇ × �B⃗ = μ0⃗J
Answer: (a)

Solution:
∇ ∙ �B⃗ = 0 Maxwell’s second equation gauss law for magnetic

field.

div curl F� = 0

14


∴ �B⃗ = ∇ × �A⃗
Such that, ∇ ∙ �B⃗ = ∇. �∇ × A�⃗� = 0

∇ × �B⃗ = μ0J̅
(from the fourth Maxwell equation)

∇ × �H�⃗ = ⃗JC + ⃗JD
∇ × �H�⃗ = ⃗J

∴ �∇ × μ�B�⃗0� = ⃗J �⃗J = ⃗JC + ⃗JD�
∇ × �B⃗ = μ0⃗J

Trick: B�⃗ is a vector, but ∇ ∙ A�⃗ is a scalar
Option (a) is correct answer

23. The cooking of the food in the microwave oven is based on the
Principle of
a) Magnetic hysteresis loss
b) Dielectric loss
c) Both magnetic hysteresis loss and dielectric loss
d) Evaporation of water
Answer: (b)
Solution
Microwave oven heats food by the process of dielectric heating.
Microwave radiation penetrates into food up to 1-2 inches and
heat the water into the food uniformly.

15


24. Plane y = 0 carries a uniform current density 30 � mA/m. At

(1, 20, -2) m, what is the magnetic field intensity?

a) −15ı̂ mA/m

b) 15ı̂ mA/m

c) 18.85ȷ̂ mA/m

d) 25ı̂ mA/m

Answer: (a)

Solution:

H��⃗ = 1 × 30k� × a� n
2

(a�n ⊥ vector to plane = ȷ� )

H = 1 × 30k� × ȷ� = −15ı�mA/m
2

25. A plane wave with an instantaneous expression for the electric

field E�(z, t) = a�xE10 sin(ωt − kz) + a�yE20 sin(ωt − kz + ϕ) is

a) Linearly polarized

b) Circularly polarized

c) Elliptically polarized

d) Horizontally polarized

Answer: (c)

Solution:

For a wave,
E�(z, t) = E10 sin(ωt − kz)a�x + a�yE20 sin(ωt − kz + ϕ)

Polarization is elliptical,
E� = a�xEx + a�yEy

16


• For E10 = E20 and ϕ = ±900, wave will be circularly
polarized

• For Ey 1800 out of phase or in phase with Ex wave is linearly
polarized

26. The force on a charge moving with velocity ⃗ under the
influence of electric and magnetic fields is given by which one
of the following?
a) q��E⃗ + �B⃗ × �v⃗�
b) q��E⃗ + �v⃗ × �H�⃗�
c) q��H�⃗ + �v⃗ × �E⃗�
d) q��E⃗ + �v⃗ × �B⃗�
Answer: (d)
Solution:
Force on charge due to electric field E�⃗
�F⃗1 = q�E⃗ ……… (1)
Force on a moving field �B⃗ i.e., Lorentz force
F�⃗2 = q��v⃗ × �B⃗� ……… (2)
So net force on the moving charge q is
F�⃗ = F�⃗1 + �F⃗2
= qE�⃗ + q��v⃗ × �B⃗�
= q��E⃗ + �v⃗ × �B⃗�

17


27. An electromagnetic field in free space (μ0,ε0) is given by:
E�⃗ = a�⃗xE0 cos(ωt − k0z) V/m

H��⃗ = a�yE0�με00 cos(ωt − k0z) A/m

Where k0 = ω �μ0ε0 . What is the average power per unit area

associated with a wave?

a) a�zE20
π

b) a�zE02
120π

c) a�zE20
240π

d) a�zE02
300π

(Given: �με00 = 120π)

Answer: (c)

Solution:

P� = E� × H� (pointing vector)

Time average of pointing vector is average power per unit area.

E� = E0e−αz cos(ωt − βz)a�⃗x

H� = E0 e−αz cos�ωt − βz − θη� a�⃗y
|η|

P�avg(z) = E02 e−2αz cos θη �a⃗z
2|η|

For given waves in free space,

η = 120π = �με00 , θη = 0, α = 0

18


�P�avg(z)� = E20 = E20
(2)(120π) 240π

P�avg(z) = E02 �a⃗z
240π

28. How much current must flow in a loop radius 1 m to produce a

magnetic field 1 mAm-1?

a) 1.0 mA

b) 1.5 mA

c) 2.0 mA

d) 2.5 mA

Answer: (c)

Solution:

Magnetic field at the center of the loop in which current-I is

flowing

H��⃗ = I �a⃗z
2r

��H�⃗� = I
2r

Given that �H��⃗� = 1 mA = 1 × 10−3A/m
m

r = 1m

∴ 1 × 10−3A = I
m 2×1

19


I = 2 × 10−3A

I = 2 mA

29. What does the expression 1 ⃗J ∙ A�⃗ represent?
2

a) Power density

b) Radiation resistance

c) Magnetic energy density

d) Electric energy density

Answer: (c)

Solution:

1 ⃗J ∙ �A⃗ = 1 A . wb (Linear current density)
2 2 m m2

1 ⃗J ∙ �A⃗ = 1 H��⃗. μH��⃗ = 1 μH2
2 2 2

Magnetic current density as

A/m → �H�⃗

�B⃗ → wb/m2 → μH��⃗

30. If �H�⃗ = �a⃗xHx − �a⃗yHy represents the H-field in a transverse

plane of an em wave travelling in the Z-direction, then what is
the �E⃗-field in the wave?
a) Z0|a�⃗xHy − �a⃗yHx|
b) Z0|a�⃗xHy + �a⃗yHx|
c) Z0|− a�⃗xHy − �a⃗yHx|
d) Z0|a�⃗xHy + �a⃗yHx|

Answer: (b)

20


Solution:

∇ × H� = ϵ0 ∂E�
∂t

a�x a�y a�z

∇ × H� = � ∂ ∂ ∂ �
∂x ∂y ∂z

Hx −Hy 0

= ∂Hy a�x + ∂Hx a�y = ϵ0 ∂E�
∂z ∂z ∂t

E� = 1 �a�x ∫ ∂t . ∂Hy + a�y ∫ ∂t . ∂Hx�
ϵ0 ∂z ∂z

= 1 �H� y a� x + H�xa�y�
ϵ0v

v = velocity = ∂z = 1
∂t �μ0ϵ0

E� = �μϵ00 �Hya�x + Hxa�y�

�μϵ00 = Z0
E� = Z0�Hya�x + Hxa�y�

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