# Electromagnetic Theory Questions and Answers – Maxwell Law 2

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Maxwell Law 2”.

1. Maxwell second equation is based on which law?
a) Ampere law
c) Lenz law
d) Coulomb law

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.

2. The Maxwell second equation that is valid in any conductor is
a) Curl(H) = Jc
b) Curl(E) = Jc
c) Curl(E) = Jd
d) Curl(H) = Jd

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.

3. In dielectric medium, the Maxwell second equation becomes
a) Curl(H) = Jd
b) Curl(H) = Jc
c) Curl(E) = Jd
d) Curl(E) = Jd

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.

4. Find the displacement current density of a material with flux density of 5sin t
a) 2.5cos t
b) 2.5sin t
c) 5cos t
d) 5sin t

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.

5. Find the conduction current density of a material with conductivity 200units and electric field 1.5 units.
a) 150
b) 30
c) 400/3
d) 300

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.
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6. Calculate the conduction density of a material with resistivity of 0.02 units and electric intensity of 12 units.
a) 300
b) 600
c) 50
d) 120

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.

7. In the conversion of line integral of H into surface integral, which theorem is used?
a) Green theorem
b) Gauss theorem
c) Stokes theorem
d) It cannot be converted

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.

8. An implication of the continuity equation of conductors is given by
a) J = σ E
b) J = E/σ
c) J = σ/E
d) J = jwEσ

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.

9. Find the equation of displacement current density in frequency domain.
a) Jd = jwεE
b) Jd = jwεH
c) Jd = wεE/j
d) Jd = jεE/w

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

10. The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field intensity.
a) 0.5i – 0.5j + 0.5k
b) 0.5i + j -1.5k
c) i – j + k
d) i + j – k

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.

11. At dc field, the displacement current density will be
a) 0
b) 1
c) Jc
d) ∞

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.

12. Both the conduction and displacement current densities coexist in which medium?
a) Only conductors in air
b) Only dielectrics in air
c) Conductors placed in any dielectric medium
d) Both the densities can never coexist

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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