The resistance is therefore 2533(731 m)(1.94 10 m)9.81 10 (2.00 10 m)(2.30 10 m)VLRiabρπ−−−Ω⋅×== == × Ωπ××Note that if b = a, then R = L/πa2= L/A, where A = πa2is the cross-sectional area of the cylinder. 35. (a) The current iis shown in Fig. 26-30 entering the truncated cone at the left end and leaving at the right. This is our choice of positive xdirection. We make the assumption that the current density Jat each value of xmay be found by taking the ratio i/Awhere A = πr2is the cone’s cross-section area at that particular value of x. The direction of GJis identical to that shown in the figure for i(our +xdirection). Using Eq. 26-11, we then find an expression for the electric field at each value of x, and next find the potential difference Vby integrating the field along the xaxis, in accordance with the ideas of Chapter 25. Finally, the resistance of the cone is given by R = V/i. Thus, JirE==π2where we must deduce how rdepends on xin order to proceed. We note that the radius
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.