problemset1

# 641 spring 2009 we assumed hz hr 0 this follows from

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Unformatted text preview: of sheet (ie, circumference of circle of radius a) � Thus, K ’s units are � |K | = d� a 2 2 3 J0 πb 2πa = Amps m � , whereas J ’s units are Amps m2 J 0 b2 3a 2 0b � K = − J3a iˆ z 4 Problem Set 1 6.641, Spring 2009 B Figure 3: A diagram of the wire with a circle C centered on the z-axis with minimum surface S (Image by MIT OpenCourseWare). Ampere’s Law � �s H · d� = C � �a J · d� + S d dt � � r �a �0 E · d� � �� � no E ﬁeld, term is 0 Choose C as a circle and S as the minimum surface that circle bounds. Now solve LHS of Ampere’s Law � C �s H · d� = 2π � 0 (Hφ iˆ ) · (rdφ)iˆ = 2πrHφ φ φ � �� � � �� � � H d� s 5 Problem Set 1 6.641, Spring 2009 We assumed Hz = Hr = 0. This follows from the symmetry of the problem. Hr = 0 because 0. In particular choose S as shown in Figure 4a. � S �a µ0 H · d� = Figure 4: A diagram of the current carrying wire of radius b with the choice for S as well as a diagram of...
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## This document was uploaded on 03/19/2014 for the course EECS 6.641 at MIT.

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