Engineering Electromagnetics with CD (McGraw-Hill Series in Electrical Engineering)

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Vector Potential Mankei Tsang Department of Electrical Engineering, California Institute of Technology, Pasadena, CA 91125 (Dated: May 19, 2005) The formalism of vector potential is extremely useful for solving the Maxwell’s equations. INTRODUCTION Let’s look at the Maxwell’s equations again: ∇ · E = ρ ǫ 0 (1) ∇ · B = 0 (2) ∇ × E = B ∂t (3) ∇ × B = μ 0 J + μ 0 ǫ 0 E ∂t (4) Imagine that you are given ρ ( r , t ) and J ( r , t ), how do you solve for E and B using the Maxwell’s eqautions? It turns out that it is not easy to do so directly, but through the formalism of vector potential one can take a detour and solve for E and B indirectly. DEFINITIONS OF VECTOR AND SCALAR POTENTIALS First look at the Gauss’s law for magnetism, ∇ · B = 0 . (5) The divergence of B is always zero, so we can always define B as the curl of another vector quantity, because the divergence of curl is always zero as well, B ≡ ∇ × A (6) ∇ · B = ∇ · ( ∇ × A ) 0 . (7)
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2 A is called the vector potential . Let’s plug the definition of vector potential, Eq. (6), into Faraday’s law, Eq. (3), ∇ × E = ∂t ( ∇ × A ) (8) ∇ × parenleftBig E + A ∂t parenrightBig = 0 (9) The curl of the quantity E + A ∂t is zero, which means that we can define it as the gradient of a scalar function, because the curl of a gradient is always zero as well, E + A ∂t ≡ −∇ V (10) ∇ × ( −∇ V ) 0 . (11) Hence E and B can be defined in terms of the vector potential A and the scalar potential V , E = −∇ V A ∂t , (12) B = ∇ × A . (13) MAXWELL’S EQUATIONS IN TERMS OF SCALAR AND VECTOR POTENTIALS These definitions, Eqs. (12) and (13), already satisfy the Gauss’s law for magnetism and Faraday’s law. Now we can plug them into Gauss’s law and Ampere’s law and see what we get. First plug Eq. (12) into the Gauss’s law, Eq. (1), ∇ · parenleftBig − ∇ V A ∂t parenrightBig = ρ ǫ 0 (14) 2 V + ∂t ( ∇ · A ) = ρ ǫ 0 . (15) We can also plug Eq. (12) and Eq. (13) into Ampere’s law, Eq. (4), ∇ × ( ∇ × A ) = μ 0 J + μ 0 ǫ 0 ∂t parenleftBig − ∇ V A ∂t parenrightBig (16) ( ∇ · A ) − ∇ 2 A = μ 0 J μ 0 ǫ 0 ∂V ∂t μ 0 ǫ 0 2 A ∂t 2 (17) 2 A μ 0 ǫ 0 2 A ∂t 2 − ∇ parenleftBig ∇ · A + μ 0 ǫ 0 ∂V ∂t parenrightBig = μ 0 J . (18)
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3 With Eqs. (15) and (18), we have successfully reduced four Maxwell’s equations into two equations, and from six unknowns (three components for E and three components for B ) to four unknowns (one scalar V and three components for A ). I will write them again together: 2 V + ∂t ( ∇ · A ) = ρ ǫ 0 (19) 2 A μ 0 ǫ 0 2 A ∂t 2 − ∇ parenleftBig ∇ · A + μ 0 ǫ 0 ∂V ∂t parenrightBig = μ 0 J . (20) But these two equations still look quite ugly and complicated. We can, however, use
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