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# 4 a we can simply add the elds of the two point

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Unformatted text preview: the wire with the choice of contour C (Image by MIT OpenCourseWare). Hz is more diﬃcult to see. It is discussed in Haus &amp; Melcher. The basic idea is to use the contour, C � (depicted in Figure 4b), to show that if Hz = 0 it would have to be nonzero even at ∞, which is not possible without sources at ∞. Now for RHS of Ampere’s Law: r&lt;b � �a J · d� = 2π � S 0 � r 0 � J0 r � ˆ � � � ˆ � iz · r dr dφiz b �� � � �� � � � d� a � J 2J0 r3 π 3b a &gt; r &gt; b = � �a J · d� = 2π � S 0 � b � 0 J0 r � ˆ iz b �� � � 2π � r � �� � �� ˆ+ · r dr dφiz 0 · iˆ · r� dr� dφiˆ z z b �0 �� � 0 2 J0 b2 π 3 Equating LHS &amp; RHS: = 2πrHφ = ⎧ ⎨ ⎩ 2 3 3b J0 r π ; 2 2 3 J0 b π ; 0; r &lt; b a&gt;r&gt;b r&gt;a 6 Problem Set 1 � H= J0 r 2 ˆ 3b iφ J 0 b2 ˆ 3r iφ ⎧ ⎨ ⎩ 6.641, Spring 2009 ; ; 0; r&lt;b a&gt;r&gt;b r&gt;a Problem 1.4 A We can simply add the ﬁelds of the two point charges. Start with the ﬁeld of a point charge q at the origin and let SR be the sphere of radius R centered at the origin. By Gauss: � � →→ −− ε0 E · d a = ρdV SR V → In this case ρ = δ (− )q , so RHS is r � ��� → ρdV = δ (− )qdxdydz = q r LHS is � − →− ε0 Er · d→ = (ε0 a SR Er )(surface area of Sr ) ���� symmetry = 4πr2 ε0 Er Equate LHS and RHS 4πr2 ε0 Er = q → − E= qˆ ir 4πr2 ε0 Convert to cartesian: Any point is given by → − = x(r, θ, φ)ˆ + y (r, θ, φ)ˆ +...
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