P27_029 - c 1 is chosen so that r = a (when x = 0);...

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29. (a) The current i is shown in Fig. 27-22 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 = πr 2 is the cone’s cross-section area at that particular value of x . The direction of ~ J is identical to that shown in the ±gure for i (our + x direction). Using Eq. 27-11, we then ±nd an expression for the electric ±eld at each value of x , and next ±nd the potential di²erence V by integrating the ±eld along the x axis, in accordance with the ideas of Chapter 25. Finally, the resistance of the cone is given by R = V/i .Thu s , J = i πr 2 = E ρ where we must deduce how r depends on x in order to proceed. We note that the radius increases linearly with x ,so(w ith c 1 and c 2 to be determined later) we may write r = c 1 + c
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Unformatted text preview: c 1 is chosen so that r = a (when x = 0); therefore, c 1 = a . Also, the coecient c 2 must be chosen so that (at the right end of the truncated cone) we have r = b (when x = L ); therefore, c 2 = ( b a ) /L . Our expression, then, becomes r = a + b a L x . Substituting this into our previous statement and solving for the eld, we nd E = i a + b a L x 2 . Consequently, the potential dierence between the faces of the cone is V = Z L E dx = i Z L a + b a L x 2 dx = i L b a a + b a L x 1 L = i L b a 1 a 1 b = i L b a b a ab = iL ab . The resistance is therefore R = V i = L ab . (b) If b = a , then R = L/a 2 = L/A , where A = a 2 is the cross-sectional area of the cylinder....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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