ch26-p035

# ch26-p035 - 35. (a) The current i is shown in Fig. 26-30...

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The resistance is therefore 2 5 33 (731 m)(1.94 10 m) 9.81 10 (2.00 10 m)(2.30 10 m) VL R ia b ρ π −− Ω⋅ × == = = × Ω π× × Note that if b = a , then R = L / π a 2 = L / A , where A = π a 2 is the cross-sectional area of the cylinder. 35. (a) The current i is shown in Fig. 26-30 entering the truncated cone at the left end and leaving at the right. This is our choice of positive x direction. We make the assumption that the current density J at each value of x may be found by taking the ratio i / A where A = π r 2 is the cone’s cross-section area at that particular value of x . The direction of G J is identical to that shown in the figure for i (our + x direction). Using Eq. 26-11, we then find an expression for the electric field at each value of x , and next find the potential difference V by integrating the field along the x axis, in accordance with the ideas of Chapter 25. Finally, the resistance of the cone is given by R = V / i . Thus, J i r E == π 2 where we must deduce how r depends on x in 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.

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