Test: Maxwell Law - 1

# Test: Maxwell Law - 1

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## 24 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Maxwell Law - 1

Test: Maxwell Law - 1 for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Maxwell Law - 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Maxwell Law - 1 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 - 1 below.
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Test: Maxwell Law - 1 - Question 1

### The first Maxwell law is based on which law?

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

Explanation: The first Maxwell equation states that Curl(E) = -dB/dt. It is based on the emf concept. Thus it is derived from the Faraday and Lenz law.

Test: Maxwell Law - 1 - Question 2

### The benefit of Maxwell equation is that

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

Explanation: The Maxwell equation relates the parameters E, D, H, B. When one parameter is known the other parameters can be easily calculated. In other words, it is used to relate an electric field parameter with its equivalent magnetic field.

Test: Maxwell Law - 1 - Question 3

### The correct sequence to find H, when D is given is

Detailed Solution for Test: Maxwell Law - 1 - Question 3

Explanation: There is no direct relation between D and H, so the option D-H is not possible. Using the formula D = εE, the parameter E can be computed from D. By Maxwell equation, Curl(E) = -dB/dt, the parameter B can be calculated. Using the formula B = μH, the parameter H can be calculated. Thus the sequence is D-E-B-H.

Test: Maxwell Law - 1 - Question 4

The curl of the electric field intensity is

Detailed Solution for Test: Maxwell Law - 1 - Question 4

Explanation: The curl of electric field intensity is Curl(E). From Maxwell law, the curl of E is a non-zero value. Thus E will be rotational.

Test: Maxwell Law - 1 - Question 5

Which of the following identities is always zero for static fields?

Detailed Solution for Test: Maxwell Law - 1 - Question 5

Explanation: The curl of gradient of a vector is always zero. This is because the gradient of V is E and the curl of E is zero for static fields.

Test: Maxwell Law - 1 - Question 6

Find the Maxwell first law value for the electric field intensity is given by A sin wt az

Detailed Solution for Test: Maxwell Law - 1 - Question 6

Explanation: The value of Maxwell first equation is Curl(E). The curl of E is zero. Thus for the given field, the value of Maxwell equation is zero. Thus the field is irrotational.

Test: Maxwell Law - 1 - Question 7

Find the electric field applied on a system with electrons having a velocity 5m/s subjected to a magnetic flux of 3.6 units.

Detailed Solution for Test: Maxwell Law - 1 - Question 7

Explanation: The electric field intensity is the product of the velocity and the magnetic flux density. Thus E = v x B, on substituting v = 5 and B = 3.6, we get E = 5 x 3.6 = 18 units.

Test: Maxwell Law - 1 - Question 8

Which of the following relations holds good?

Detailed Solution for Test: Maxwell Law - 1 - Question 8

Explanation: The force of a electrostatic field in given by F = Eq. The force on a conductor is given by F = BIL. In the case when a charge exists on a conductor, both the forces can be equated. Thus Eq = BIL is true.

Test: Maxwell Law - 1 - Question 9

When the Maxwell equation is expressed in frequency domain, then which substitution is possible?

Detailed Solution for Test: Maxwell Law - 1 - Question 9

Explanation: The conversion of time to frequency domain in Maxwell equation is given by the Fourier Transform. Differentiation in time gives jw in frequency domain. Thus d/dt = jw in frequency domain.

Test: Maxwell Law - 1 - Question 10

Calculate the emf of a material having a flux linkage of 2t2 at time t = 1second.

Detailed Solution for Test: Maxwell Law - 1 - Question 10

Explanation: The emf of a material is given by Vemf = -dλ/dt. On substituting λ = 2t2, the emf is 4t. At t = 1 sec, the emf will be 4 units.

Test: Maxwell Law - 1 - Question 11

Calculate the emf of a material having flux density 5sin t in an area of 0.5 units.

Detailed Solution for Test: Maxwell Law - 1 - Question 11

Explanation: The emf can be written as Vemf = -d(∫B.ds)/dt. It can be written as Vemf = -B= -5sin t, since the integration and differentiation gets cancelled.

Test: Maxwell Law - 1 - Question 12

To find D from B, sequence followed will be

Detailed Solution for Test: Maxwell Law - 1 - Question 12

Explanation: Using Maxwell equation, from B we can calculate E by Curl(E) = -dB /dt. From E, D can be calculated by D = εE. Thus the sequence is B->E->D.

Test: Maxwell Law - 1 - Question 13

Maxwell second equation is based on which law?

Detailed Solution for Test: Maxwell Law - 1 - Question 13

Explanation: The second Maxwell equation is based on Ampere law. It states that the field intensity of a system is same as the current enclosed by it, i.e, Curl(H) = J.

Test: Maxwell Law - 1 - Question 14

The Maxwell second equation that is valid in any conductor is

Detailed Solution for Test: Maxwell Law - 1 - Question 14

Explanation: For conductors, the conductivity parameter σ is significant and only the conduction current density exists. Thus the component J = Jc and Curl(H) = Jc.

Test: Maxwell Law - 1 - Question 15

In dielectric medium, the Maxwell second equation becomes

Detailed Solution for Test: Maxwell Law - 1 - Question 15

Explanation: In dielectric medium conductivity σ will be zero. So the current density has only the displacement current density. Thus the Maxwell equation will be Curl(H) = Jd.

Test: Maxwell Law - 1 - Question 16

Find the displacement current density of a material with flux density of 5sin t

Detailed Solution for Test: Maxwell Law - 1 - Question 16

Explanation: The displacement current density is the derivative of the flux density. Thus Jd = dD/dt. Put D = 5sin t in the equation, we get Jd = 5cos t units.

Test: Maxwell Law - 1 - Question 17

Find the conduction current density of a material with conductivity 200units and electric field 1.5 units.

Detailed Solution for Test: Maxwell Law - 1 - Question 17

Explanation: The conduction current density is given by Jc = σE, where σ = 200 and E = 1.5. Thus we get, Jc = 200 x 1.5 = 300 units.

Test: Maxwell Law - 1 - Question 18

Calculate the conduction density of a material with resistivity of 0.02 units and electric intensity of 12 units.

Detailed Solution for Test: Maxwell Law - 1 - Question 18

Explanation: The conduction density is given by Jc = σE, where σ is the inverse of resistivity and it is 1/0.02 = 50. Thus we get, Jc = 50 x 12 = 600 units.

Test: Maxwell Law - 1 - Question 19

In the conversion of line integral of H into surface integral, which theorem is used?

Detailed Solution for Test: Maxwell Law - 1 - Question 19

Explanation: To convert line integral to surface integral, i.e, in this case from line integral of H to surface integral of J, we use the Stokes theorem. Thus the Maxwell second equation can be written as ∫H.dl = ∫∫J.ds

Test: Maxwell Law - 1 - Question 20

An implication of the continuity equation of conductors is given by

Detailed Solution for Test: Maxwell Law - 1 - Question 20

Explanation: The continuity equation indicates the current density in conductors. This is the product of the conductivity of the conductor and the electric field subjected to it. Thus J = σE is the implication of the continuity equation for conductors.

Test: Maxwell Law - 1 - Question 21

Find the equation of displacement current density in frequency domain.

Detailed Solution for Test: Maxwell Law - 1 - Question 21

Explanation: The displacement current density is Jd = dD/dt. Since D = εE and in frequency domain d/dt = jw, thus we get Jd = jwεE.

Test: Maxwell Law - 1 - Question 22

The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field intensity.

Detailed Solution for Test: Maxwell Law - 1 - Question 22

Explanation: By Maxwell second equation, the curl of H is same as the sum of conduction current density and displacement current density. Thus Curl(H) = J = 0.5i + j – 1.5k units.

Test: Maxwell Law - 1 - Question 23

At dc field, the displacement current density will be

Detailed Solution for Test: Maxwell Law - 1 - Question 23

Explanation: The DC field refers to zero frequency. The conduction current is independent of the frequency, whereas the displacement current density is dependent on the frequency, i.e, Jd = jwεE. Thus at DC field, the displacement current density will be zero.

Test: Maxwell Law - 1 - Question 24

Both the conduction and displacement current densities coexist in which medium?

Detailed Solution for Test: Maxwell Law - 1 - Question 24

Explanation: Conduction density exists only for good conductors and displacement density is for dielectrics in any medium at high frequency. Thus both coexist when a conductor is placed in a dielectric medium.

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests