InterviewSolution
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InterviewSolution
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A long round conductor of cross- sectional area S is made of material whose resistivity depends only on distance r from the axis of the conductor as rho = (alpha)/(r^2) , where alpha is a constant. Find (a) the resistance per unit length of such a conductor. (b) the electric field strength in the conductor due to which a current i flows through it . |
Answer» Solution :Consider a CYLINDRICAL element of radii between r and (r+ DR) . Its resistance  `dR = (rhol)/(2pi dr) ` (or)` 1/(dR) = (2pi r dr)/(rhol)""……(i)` `THEREFORE 1/R = int_0^a 1/(dR) = int_0^a (2pi)/(rhol)r dr` (where a is the radius of the conductor ) `int_0^a (2pir dr)/(((alpha)/(r^2))l) = (2pi)/(alphal) int_0^a r^3 dr = (2pi)/(alpha l) ((a^4)/(4)) = ((pia^2)^2)/(2pialphal) = (S^2)/(2pi alpha l)` `R = (2pi alpha l)/(S^2) "".....(ii)` The resistance per unit length of wire `R= (2pi alpha)/(S^2)` (b) Equation (ii) can be written as `R= ((2pialpha)/(S)) ((l)/(S))` COMPARE with `R = (rhol)/(S) , ` we get `rho = (2pi alpha)/(S)` By Ohm.s law `E= jrho = i/S xx (2pi alpha)/(S) = (2pi alpha i)/(S^2)` |
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