FWLec14 - A. F. Peterson: Notes on Electromagnetic Fields &...

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Fields & Waves Note #14 Ampere’s Circuital Law in Differential Form Objectives: Provide an example where Ampere’s Law is used to determine the magnetic field due to a distribution of volume current density. Then, use the integral form of Ampere’s Law to obtain the differential form of Ampere’s Law. Discuss the curl operator that arises in the differential form, and illustrate its physical significance. The integral form of Ampere’s Law In this Note, we continue exploring Ampere’s Circuital Law, introduced in Note #13: Hd Jd S CS = ÚÚ Ú l (14.1) We begin by using Ampere’s Law to determine the magnetic field due to a given distribution of volume current density. Example: A uniform current density J I a z = p 2 ˆ (14.2) exists within the region r < a . Find the magnetic field produced by this current in the regions < a and > a . Solution: The cylindrical symmetry of the geometry is similar to the example in Note #13, and again we expect that the H -field will not be a function of either f or z , and will have only a component. We choose the contour C to be a circle in a constant- z plane, centered at the z -axis (Figure 1). For such a circle of any general radius , H d H C = = = l ˆ fr f pr 0 2 2 (14.3) As in the preceding example, the surface associated with this contour can be chosen to be the disk of radius , in a constant- z plane, and the dS vector is obtained as
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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 dS d d z = rrf ˆ (14.4) For r < a , the total current through S is Jd S I a zd d z a S = Ê Ë Á ˆ ¯ ˜ < ÚÚ Ú Ú = p f 2 0 2 =0 ˆˆ (14.5) Equation (14.5) can be evaluated to yield S I a d I a I a a S = =< ÚÚ Ú 2 2 2 2 2 2 2 2 rr = (14.6) For > a , a different result is obtained (due to the fact that the current is limited to the region within radius a) S I a d z a S a = Ê Ë Á ˆ ¯ ˜ > ÚÚ Ú Ú = 2 0 2 (14.7) Equation (14.7) can be evaluated to yield S I a S => ÚÚ (14.8) Therefore, by equating (14.3) with (14.7) and (14.8), we obtain H I a a I a pr = < > Ï Ì Ô Ô Ó Ô Ô 2 2 2 (14.9) Since the mathematical form of the current is different in the two regions, the expression for magnetic field is also different in each region. Outside the source region, the magnetic field is identical to that of a thin wire carrying current I.
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This note was uploaded on 01/27/2011 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Tech.

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FWLec14 - A. F. Peterson: Notes on Electromagnetic Fields &...

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