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# ch26_p33 - 33(a The current i is shown in Fig 26-29...

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33. (a) The current i is shown in Fig. 26-29 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 ρ where we must deduce how r depends on x in order to proceed. We note that the radius increases linearly with x , so (with c 1 and c 2 to be determined later) we may write r c c x = + 1 2 . Choosing the origin at the left end of the truncated cone, the coefficient c 1 is chosen so that r = a (when x = 0); therefore,

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ch26_p33 - 33(a The current i is shown in Fig 26-29...

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