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Two longparallelwiresof negligibleresistance are connectedat one endto a reasitanceR and at the otherend to adcvoltagesource. The distancebetweenthe axesof the wires is eta = 20 timesgreaterthan the cross-sectionalradiusof each wire. At whatvalueof resistanceR doesthe resultantforceof interaction betweenthe wiresturn into zero ? |
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Answer» Solution :There are EXCESS surface chargeson eachwire (irrespectiveof whether the current is flowingthroughthen or net). Hencein addtiionto the magneticforce `vec(F)_(m)`, we must takeinto accountthe ELECTRICFORCE `vec(F)_(e)`. Suppose that an excesschagre `lambda` correspondsto a unitlength of the wire, then electricforceexertedper unitlength of the wireby otherwire can befound with the help of Gauss's theorem. `F_(e) = lambda E = lambda (1)/(4pi epsilon_(0)) (2 lambda)/(l) = (2 lambda^(2))/(4pi epsilon_(0) l)` ....(1) where `l` is the distance between the area of the wires. The magentic forceacting per unitlength of the wirecan be found with the helpof the theroem on circularion of vector `vec(B)` `F_(m) = (mu_(0))/(4pi) (2i^(2))/(l)`, where `i` is the current in the wire. ..(2) Now, from the relation, `lambda = C varphi`, where `C` is the capacitanceof the wires per unit lengthsand is given in problem3.108 and `varphi = iR` `lambda = (pi epsilon_(0))'/(ln ETA) iR` or, `(i)/(lambda) (ln eta)/(pi epsilon_(0)R)` .....(3) Dividing(2) by (1) and then substuting the valuesof `(i)/(lambda)` from (3) we get, `(F_(m))/(F_(e)) = (mu_(0))/(epsilon_(0)) ((ln eta)^(2))/(pi^(2) R^(2))` The resultantforceof interaction vanisheswhen this ratiounity. This is possible when `R = R_(0)`, where `R_(0) = SQRT((mu_(0))/(epsilon_(0))) (ln eta)/(pi) = 0.36 k OMEGA` 
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