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The magnetic vector potential for a line current will be inversely proportional to
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.
d
See lessExplanation: 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.
The Laplacian of the magnetic vector potential will be
a 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.
a
See lessExplanation: 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.
Find the flux density B when the potential is given by x i + y j + z k in air.
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.
b
See lessExplanation: 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.
Find the vector potential when the field intensity 60×2 varies from (0,0,0) to (1,0,0).
b 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.
b
See lessExplanation: The field intensity H = -Grad(V). To get V, integrate H with respect to the variable. Thus V = -∫H.dl = -∫60×2 dx = -20×3 as x = 0->1 to get -20.
Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.
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.
c
See lessExplanation: 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.
The value of ∫ H.dL will be
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.
b
See lessExplanation: 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.
Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.
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.
b
See lessExplanation: The magnetic field intensity is given by H = -Grad(Vm). The gradient of Vm is 1 + 2 + 3 = 6. Thus H = -6 units.
The magnetic vector potential is a scalar quantity.
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.
b
See lessExplanation: 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.
Find current density J when B = 50 x 10-6 units and area dS is 4 units.
a Explanation: To get H, H = B/μ = 50 x 10-6/ 4π x 10-7 = 39.78 units. Also H = ∫ J.dS, where H = 39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.
a
See lessExplanation: To get H, H = B/μ = 50 x 10-6/ 4π x 10-7 = 39.78 units. Also H = ∫ J.dS,
where H = 39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.
Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.
b Explanation: We know that Curl (E) = -dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B, integrate the -curl(E) with respect to time to get B = -∫(4sin z i – ex j – 3cos yk)dt.
b
See lessExplanation: We know that Curl (E) = -dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B, integrate the -curl(E) with respect to time to get B = -∫(4sin z i – ex j – 3cos yk)dt.