325_Sp2011_10_Scalar_and_vector_potential part1

# 325_Sp2011_10_Scalar_and_vector_potential part1 - 10....

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 1 10. Scalar and Vector potentials for magnetic field EE325 Mikhail Belkin

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 2 Potential functions for magnetic fields recall that in electrostatics it was sometimes easier to find the electrostatic potential, then get the electric field from the gradient of the potential we could either integrate over charge density to get V, or solve a differential equation (Poisson or Laplace) subject to boundary conditions is there a potential function for H, and is it easier to find?? let’s assume there is a scalar magnetic potential V m that gives H from the negative gradient (just like in electrostatics) what conditions, if any, are there on V m ? m H V = -∇ r
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 3 Potential functions for magnetic fields assume there is a scalar magnetic potential V m that gives H from the negative gradient if there is current present, we already know so what is the curl of the gradient? m H V = -∇ r H J ∇× = r r ( 29 m V J ∇× -∇ = r ˆ ˆ ˆ m m m m V V V V x y z x y z -∇ = - + + ÷ r ˆ ˆ ˆ y y x x z z F F F F F F F x y z y z z x x y ∇× = - + - + - ÷ ÷ ÷ r

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 4 Potential functions for magnetic fields what is the curl of the gradient? in general, for any function f, the curl of the gradient is identically zero 2 2 2 2 2 2 ˆ ˆ ˆ 0 m m m m m m V V V V V V x y z y z y z z x z x x y x y = - - + - + - = ÷ ÷ ÷ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ( 29 ˆ ˆ ˆ m m m m m m m V V V V V V y y z x z x V x y z y z z x x y ÷ ÷ ÷ ÷ ÷ ∇× -∇ = - - + - + - ÷ ÷ ÷ ÷ ÷ ÷ ( 29 0 m V ∇× -∇ = ( 29 0 f ∇× ∇ =
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 5 Back to our scalar magnetic potential so far we have so it only makes sense in current free regions of space we also have from the divergence in current free regions Laplace’s equation is satisfied m H V = -∇ r H J ∇× = r r ( 29 m V J ∇× -∇ = r ( 29 0 m V ∇× -∇ = 0 m H V if J = -∇ = r r 0 B H μ = ∇ = r r r r g g ( 29 0 m V μ∇ -∇ = r g ( 29 2 0 m m V V ∇ ∇ = ∇ = r g

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 6
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## 325_Sp2011_10_Scalar_and_vector_potential part1 - 10....

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