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Two long parallel wiresare locatedin a poorly conductingmedium with respectivityrho. The distance betweenthe axesof the wires is equal to l, the cross -section radiusof eachwire equals a. In the casea lt lt l find, (a) the current density at the point equally removed from the axesof the wireby a distancerif the potentialdifferencebetween teh wires is equal to V, (b) the electric resistancesof the medium per unitlength of the wires. |
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Answer» Solution :(a) The wires themselves will be ASSUMED to be perfect conductors so the RESISTANCE is entirelydue to the medium. If the wireis of length `L`,the resistance `R` of the medium is `alpha (1)/(L)` becausedifferent sections of the wire are CONNECTED in parallel(by the medium) ratherthan in series. Thus if `R_(1)` is the resistance per unit length of the wirethan `R = R_(1)//L`. Unit of `R_(1)` is obnm-meter. The potential at a point `P` is by symmetry and suuperposition (for `l gt gt a`) `varphi = (A)/(2) In (r_(1))/(a) - (A)/(2) In (r_(2))/(a)` `= (A)/(2) In (r_(1))/(r_(2))` Then `varphi_(1) = (V)/(2) = (A)/(2) In (a)/(l)` (for the POTENTIALOF 1) or, `A = - V//IN (l)/(a)` and`varphi = - (V)/(2 In l//a) In r_(1)//r_(2)` We then calculate the field at a point `P` which is equidistanat from 1 & 2 and at a distance `r` from both : Then`E = (V)/(2In l//a) ((1)/(r)) xx 2 sin THETA` `= (Vl)/(2 In l//a) (1)/(r^(2))` and`J = sigma E = (1)/(rho) (V)/(2In l//a) (1)/(r^(2))` (b) Near either wire `E = (V)/(2 In l//a) (1)/(a)` and`J = sigma E = (1)/(rho) (V)/(2 In l//a)` Then `I = (V)/(R) = L (V)/(R_(1)) = J 2pi a L` which gives `R_(1) = (rho)/(pi) In l//a` 
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