Test: Maxwell Law - 1


24 Questions MCQ Test Electromagnetic Theory | Test: Maxwell Law - 1


Description
This mock test of Test: Maxwell Law - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 24 Multiple Choice Questions for Electrical Engineering (EE) Test: Maxwell Law - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Maxwell Law - 1 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Maxwell Law - 1 exercise for a better result in the exam. You can find other Test: Maxwell Law - 1 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

The first Maxwell law is based on which law?

Solution:

Answer: d
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.

QUESTION: 2

The benefit of Maxwell equation is that

Solution:

Answer: a
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.

QUESTION: 3

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

Solution:

Answer: a
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.

QUESTION: 4

The curl of the electric field intensity is

Solution:

Answer: b
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.

QUESTION: 5

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

Solution:

Answer: d
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.

QUESTION: 6

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

Solution:

Answer: a
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.

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.

Solution:

Answer: b
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.

QUESTION: 8

Which of the following relations holds good?

Solution:

Answer: c
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.

QUESTION: 9

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

Solution:

Answer: c
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.

QUESTION: 10

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

Solution:

Answer: b
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.

QUESTION: 11

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

Solution:

Answer: d
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.

QUESTION: 12

To find D from B, sequence followed will be

Solution:

Answer: a
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.

QUESTION: 13

Maxwell second equation is based on which law?

Solution:

Answer: a
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.

QUESTION: 14

The Maxwell second equation that is valid in any conductor is

Solution:

Answer: a
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.

QUESTION: 15

In dielectric medium, the Maxwell second equation becomes

Solution:

Answer: a
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.

QUESTION: 16

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

Solution:

Answer: c
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.

QUESTION: 17

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

Solution:

Answer: d
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.

QUESTION: 18

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

Solution:

Answer: b
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.

QUESTION: 19

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

Solution:

Answer: c
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

QUESTION: 20

An implication of the continuity equation of conductors is given by

Solution:

Answer: a
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.

QUESTION: 21

Find the equation of displacement current density in frequency domain.

Solution:

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

QUESTION: 22

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

Solution:

Answer: b
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.

QUESTION: 23

At dc field, the displacement current density will be

Solution:

Answer: a
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.

QUESTION: 24

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

Solution:

Answer: c
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.

Similar Content

Related tests