This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Magnetic Vector Potential”.
1. The magnetic vector potential is a scalar quantity.
Explanation: The magnetic vector potential could be learnt as a scalar. But it is actually a vector quantity, which means it has both magnitude and direction.
2. Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.
Explanation: The magnetic field intensity is given by H = -Grad(Vm). The gradient of Vm is 1 + 2 + 3 = 6. Thus H = -6 units.
3. The value of ∫ H.dL will be
Explanation: By Stoke’s theorem, ∫ H.dL = ∫ Curl(H).dS and from Ampere’s law, Curl(H) = J. Thus ∫ H.dL = ∫ J.dS which is nothing but current I.
4. Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.
Explanation: The field intensity is given by H = – Grad(V). The gradient is given by 0 – 12cos y. At the origin, the gradient will be -12 cos 0 = -12. Thus the field intensity will be 12 units.
5. Find the vector potential when the field intensity 60x2 varies from (0,0,0) to (1,0,0).
Explanation: The field intensity H = -Grad(V). To get V, integrate H with respect to the variable. Thus V = -∫H.dl = -∫60x2 dx = -20x3 as x = 0->1 to get -20.
6. Find the flux density B when the potential is given by x i + y j + z k in air.
a) 12π x 10-7
b) -12π x 10-7
c) 6π x 10-7
d) -6π x 10-7
Explanation: The field intensity H = -Grad(V). Since the given potential is a position vector, the gradient will be 3 and H = -3. Thus the flux density B = μH = 4π x 10-7 x (-3) = -12π x 10-7 units.
7. The Laplacian of the magnetic vector potential will be
a) –μ J
b) – μ I
c) –μ B
d) –μ H
Explanation: The Laplacian of the magnetic vector potential is given by Del2(A) = -μ J, where μ is the permeability and J is the current density.
8. The magnetic vector potential for a line current will be inversely proportional to
Explanation: The magnetic vector potential for the line integral will be A = ∫ μIdL/4πR. It is clear that the potential is inversely proportional to the distance or radius R.
9. The current element of the magnetic vector potential for a surface current will be
a) J dS
b) I dL
c) K dS
d) J dV
Explanation: The magnetic vector potential for the surface integral is given by A = ∫ μKdS/4πR. It is clear that the current element is K dS.
10. The relation between flux density and vector potential is
a) B = Curl(A)
b) A = Curl(B)
c) B = Div(A)
d) A = Div(B)
Explanation: The magnetic flux density B can be expressed as the space derivative of the magnetic vector potential A. Thus B = Curl(A).
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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