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A coll of radius r = 25 cm wound of a thin a copper wireof lengthl = 500 mrotates with an angularvelocityomega = 300 rad//sabout itsaxis. The coll is connectedtoa baliistic galvanometerby meansof slidingcontacts. Thetotal resistanace of thecircuitis equalt R = 21 Omega. Find teh specific totalresistanceof the circuit is equalto R = 21 Omega. Find the specfiic chargeof currentcarriesin copperif a sudden stoppageof thecoil makesa charge q = 10 nC flow through the galvanometer. |
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Answer» Solution :In a rotating route, to first order in `omega`, the main effect is a colotis froce `2 m vec(v) xx vec(omega)`. This unbalanced froce will cause electronsto REACT by setting up a magnetic field `vec(B)` so that the magnitude force `e vec(v) xx vec(B)` balances the coritolis force. Thus `- (epsilon)/(2M) vec(B) -vec(omega)` or,`vec(B) = - (2m)/(epsilon) vec(omega)` The flux assocatied with this is `Phi = N pi r^(2) B = N pi r^(2) (2m)/(e) omega` where `N = (l)/(2 pi r)` is the number of thrus of the RING. If `omega` CHANGES(and then is time for electron to rearrange) then `B` ALSO chargesand so does`Phi`. An `emf` will beinduced and a currentwill flow. This is `I = N pi r^(2) (2m)/(e) Phi//R` The total charge flowingthrough the ballastic galvanometer as the ringis stopped, is `q = N pi r^(2)// (2m)/(e) omega//R` So,`(e)/(m) = (2N pi r^(2) omega)/(q R) = (l omega r)/(q R)` |
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