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546    CHAPTER 15  Special Functions and Eigenfunction Expansions

                                 15.3.7 Alternating Current and the Skin Effect
                                 We will use a modified Bessel function to analyze the flow of alternating current in a wire of
                                 circular cross section. Begin with general principles named for Ampére and Faraday. Ampére’s
                                 law states that the integral of the magnetic force around a circuit is equal to 4π times the integral
                                 of the current through the circuit. Faraday’s law states that the integral of the electric force around
                                 a circuit equals the negative of the time derivative of the magnetic induction through the circuit.
                                    We will determine the current density at radius r in a wire of radius a.Let ρ be the specific
                                 resistance of the wire, μ its permeability, x(r,t) the current density, and H(r,t) the magnetic
                                 intensity at radius r and time t. Apply Ampére’s law to a circle of radius r having its axis along
                                 the axis of the wire:
                                                                     	  r
                                                           2πrH = 4π   2πrx(ξ,t)dξ,
                                                                      0
                                 or
                                                                     	  r
                                                             rH = 4π   rx(ξ,t)dξ.                      (15.14)
                                                                      0
                                 Then
                                                                 ∂
                                                                  (rH) = 4πxr,
                                                                ∂r
                                 hence
                                                              1 ∂
                                                                  (rH) = 4πx(r,t).                     (15.15)
                                                              r ∂r
                                 Now apply Faraday’s law to the rectangular circuit having one side of length L along the axis of
                                 the wire, and the other side of length r. We get
                                                                          ∂  	  r
                                                    ρLx(0,t) − ρLx(r,t) =−    μLH(ξ,t)dξ.
                                                                         ∂t  0
                                 Differentiate this equation with respect to r to obtain
                                                                   ∂x    ∂ H
                                                                 ρ   = μ    .                          (15.16)
                                                                   ∂r    ∂t
                                 Use equations (15.15) and (15.16) to eliminate H. First multiply (15.16) by r to obtain
                                                                   ∂x     ∂ H
                                                                ρr   = μr    .
                                                                   ∂r     ∂t
                                 Differentiate this equation with respect to r:

                                                     ∂    ∂x      ∂   ∂ H      ∂   ∂
                                                   ρ     r    = μ    r     = μ       (rH)
                                                     ∂r   ∂r     ∂r    ∂t      ∂t  ∂r
                                                                  ∂             ∂x
                                                              = μ  (4πxr) = 4πμr   ,
                                                                 ∂t              ∂t
                                 in which we substituted from equation (15.15) at the last step. Now we have

                                                               ∂   ∂x         ∂x
                                                             ρ    r    = 4πμr   .                      (15.17)
                                                              ∂r   ∂r         ∂t
                                 The strategy is to solve this partial differential equation for x(r,t), then obtain H(r,t) from
                                 equation (15.14). To do this, assume that the alternating current flowing through the wire has
                                 period 2π/ω and is given by C cos(ωt), with C constant. It is convenient to introduce complex
                                 quantities and write
                                                             z(r,t) = x(r,t) + iy(r,t).




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                                   October 14, 2010  15:20  THM/NEIL   Page-546        27410_15_ch15_p505-562
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