Substituting the values given, we obtain 423(30.0m)(7.80 10 A)0.70 m5.22 10 A2 (4.00 10)(35.0 m)(35.0 m0.70 m)iπ−Ω⋅×==××Ω+. 34. We follow the procedure used in Sample Problem 26-5. Since the current spreads uniformly over the hemisphere, the current density at any given radius rfrom the striking point is 2/2JI r=. From Eq. 26-10, the magnitude of the electric field at a radial distance ris 22wwIEJrρ, where 30mw=Ω⋅
This is the end of the preview. Sign up
access the rest of the document.