Test: Maxwell Law - 2


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20 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Maxwell Law - 2

Test: Maxwell Law - 2 for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Maxwell Law - 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Maxwell Law - 2 MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Maxwell Law - 2 below.
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Test: Maxwell Law - 2 - Question 1

The charge density of a electrostatic field is given by

Detailed Solution for Test: Maxwell Law - 2 - Question 1

Answer: d
Explanation: From the Gauss law for electric field, the volume charge density is the divergence of the electric flux density of the field. Thus Div(D) = ρv.

Test: Maxwell Law - 2 - Question 2

In the medium of free space, the divergence of the electric flux density will be

Detailed Solution for Test: Maxwell Law - 2 - Question 2

Answer: b
Explanation: In free space or air, the charge density will be zero. In other words, the conduction is possible in mere air medium. By gauss law, since the charge density is same as the divergence of D, the Div(D) in air/free space will be zero.

Test: Maxwell Law - 2 - Question 3

In a medium other than air, the electric flux density will be

Detailed Solution for Test: Maxwell Law - 2 - Question 3

Answer: d
Explanation: In any medium other than the air, the conduction is possible, due to the charge carriers. Thus charge density is also non-zero. We can write from Gauss law that Div(D) is non-zero. When the divergence is said to be non-zero, the field is not solenoidal or called as divergent field.

Test: Maxwell Law - 2 - Question 4

For a solenoidal field, the surface integral of D will be,

Detailed Solution for Test: Maxwell Law - 2 - Question 4

Answer: a
Explanation: For a solenoidal field, the divergence will be zero. By divergence theorem, the surface integral of D and the volume integral of Div(D) is same. So as the Div(D) is zero for a solenoidal field, the surface integral of D is also zero.

Test: Maxwell Law - 2 - Question 5

In a dipole, the Gauss theorem value will be

Detailed Solution for Test: Maxwell Law - 2 - Question 5

Answer: b
Explanation: The Gauss theorem for an electric field is given by Div(D)= ρ. In a dipole only static charge exists and the divergence will be zero. Thus the Gauss theorem value for the dipole will be zero.

Test: Maxwell Law - 2 - Question 6

Find the electric flux density of a material whose charge density is given by 12 units in a volume region of 0.5 units.

Detailed Solution for Test: Maxwell Law - 2 - Question 6

Answer: c
Explanation: By Gauss law, Div(D) = ρv. To get D, integrate the charge density given. Thus D = ∫ρv dv, where ρv = 12 and ∫dv = 0.5. We get, D = 12 x 0.5 = 6 units.

Test: Maxwell Law - 2 - Question 7

From the Gauss law for electric field, we can compute which of the following parameters?

Detailed Solution for Test: Maxwell Law - 2 - Question 7

Answer: c
Explanation: From the Gauss law for electric field, we can find the electric flux density directly. On substituting, D= ε E, the electric field intensity can be calculated.

Test: Maxwell Law - 2 - Question 8

The charge density of a system with the position vector as electric flux density is

Detailed Solution for Test: Maxwell Law - 2 - Question 8

Answer: d
Explanation: The divergence of the electric flux density is the charge density. For a position vector xi + yj + zk, the divergence will be 1 + 1 + 1 = 3. Thus by Gauss law, the charge density is also 3.

Test: Maxwell Law - 2 - Question 9

The sequence for finding E when charge density is given is

Detailed Solution for Test: Maxwell Law - 2 - Question 9

Answer: a
Explanation: From the given charge density ρv, we can compute the electric flux density by Gauss law. Since, D = εE, the electric field intensity can also be computed. Thus the sequence is E-D-ρv.

Test: Maxwell Law - 2 - Question 10

The Gauss law employs which theorem for the calculation of charge density?

Detailed Solution for Test: Maxwell Law - 2 - Question 10

Answer: c
Explanation: The Gauss divergence theorem is given by ∫ D.ds = ∫Div(D).dv. From the theorem value, we can compute the charge density. Thus Gauss law employs the Gauss divergence theorem.

Test: Maxwell Law - 2 - Question 11

Which quantity is solenoidal in the electromagnetic theory?

Detailed Solution for Test: Maxwell Law - 2 - Question 11

Answer: d
Explanation: The divergence of the magnetic flux density is zero. This is the Maxwell fourth equation. As the divergence is zero, the quantity will be solenoidal or divergent less.

Test: Maxwell Law - 2 - Question 12

Which equation will be true, if the medium is considered to be air?

Detailed Solution for Test: Maxwell Law - 2 - Question 12

Answer: b
Explanation: From the Gauss law for magnetic field, the divergence of the magnetic flux density is zero. Also B = μH. Thus divergence of H is also zero, i.e, Div(H) = 0 is true.

Test: Maxwell Law - 2 - Question 13

Find the sequence to find B when E is given.

Detailed Solution for Test: Maxwell Law - 2 - Question 13

Answer: a
Explanation: From E, D can be computed as D = εE. Using the Ampere law, H can be computed from D. Finally, B can be calculated from H by B = μH.

Test: Maxwell Law - 2 - Question 14

The Gauss law for magnetic field is valid in

Detailed Solution for Test: Maxwell Law - 2 - Question 14

Answer: d
Explanation: The Gauss law for magnetic field states that the divergence of B is always zero. This is valid for all cases like free space, dielectric medium etc.

Test: Maxwell Law - 2 - Question 15

The sequence for finding H from E is

Detailed Solution for Test: Maxwell Law - 2 - Question 15

Answer: a
Explanation: From E, we can compute B using the Maxwell first law. Using B, the parameter H can be found since B = μH. Thus the sequence is E-B-H is true.

Test: Maxwell Law - 2 - Question 16

The reason for non existence of magnetic monopoles is

Detailed Solution for Test: Maxwell Law - 2 - Question 16

Answer: a
Explanation: Practically monopoles do not exist, due to the connection between north and south poles. But theoretically, they exist. The reason for their non- existence practically is that, the magnetic field confined to two poles cannot be split or confined to a single pole.

Test: Maxwell Law - 2 - Question 17

The non existence of the magnetic monopole is due to which operation?

Detailed Solution for Test: Maxwell Law - 2 - Question 17

Answer: b
Explanation: The Maxwell fourth law or the Gauss law for magnetic field states that the divergence of B is zero, implies the non existence of magnetic monopoles. Thus the operation involved is divergence.

Test: Maxwell Law - 2 - Question 18

Will dielectric breakdown lead to formation of magnetic monopole? 

Detailed Solution for Test: Maxwell Law - 2 - Question 18

Answer: b
Explanation: When dielectric breakdown occurs, the material loses its dielectric property and becomes a conductor. When it is subjected to a magnetic field, north and south flux lines coexists, giving magnetic force. Thus there exists magnetic dipole. Suppose if the conductor is broken into very small pieces, still there exist a magnetic dipole in every broken part. In other words, when a piece is broken into half, there cannot exist a north pole in one half and a south pole in the other. Thus monopoles never exist.

Test: Maxwell Law - 2 - Question 19

Which equation will hold good for a magnetic material?

Detailed Solution for Test: Maxwell Law - 2 - Question 19

Answer: d
Explanation: We know that the divergence of B is zero. From Stokes theorem, the surface integral of B is equal to the volume integral of divergence of B. Thus surface integral of B is also zero.

Test: Maxwell Law - 2 - Question 20

The dipole formation in a magnet is due to

Detailed Solution for Test: Maxwell Law - 2 - Question 20

Answer: a
Explanation: In any magnetic material or magnet, the dipoles exist. This is due to the magnetic lines of force joining the north to south poles. The interaction between these two poles together leads to dipole formation.

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