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QUESTION: 1

The magnetic vector potential is a scalar quantity.

Solution:

Answer: b

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.

QUESTION: 2

Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.

Solution:

Answer: b

Explanation: The magnetic field intensity is given by H = -Grad(Vm). The gradient of Vm is 1 + 2 + 3 = 6. Thus H = -6 units.

QUESTION: 3

The value of ∫ H.dL will be

Solution:

Answer: b

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.

QUESTION: 4

Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.

Solution:

Answer: c

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.

QUESTION: 5

Find the vector potential when the field intensity 60x^{2} varies from (0,0,0) to (1,0,0).

Solution:

Answer: b

Explanation: The field intensity H = -Grad(V). To get V, integrate H with respect to the variable. Thus V = -∫H.dl = -∫60x^{2} dx = -20x^{3} as x = 0->1 to get -20.

QUESTION: 6

Find the flux density B when the potential is given by x i + y j + z k in air.

Solution:

Answer: b

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.

QUESTION: 7

The Laplacian of the magnetic vector potential will be

Solution:

Answer: a

Explanation: The Laplacian of the magnetic vector potential is given by Del^{2}(A) = -μ J, where μ is the permeability and J is the current density.

QUESTION: 8

The magnetic vector potential for a line current will be inversely proportional to

Solution:

Answer: d

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.

QUESTION: 9

The current element of the magnetic vector potential for a surface current will be

Solution:

Answer: c

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.

QUESTION: 10

The relation between flux density and vector potential is

Solution:

Answer: a

Explanation: The magnetic flux density B can be expressed as the space derivative of the magnetic vector potential A. Thus B = Curl(A).

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