12.07.2015 Views

The thorny way of truth - Free Energy Community

The thorny way of truth - Free Energy Community

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1C8 -However, if there are Ni and N2 free elements per unit volume andcross sectional areas are A iand A2|Q,|-N,A,|Oe||ds,| , |02|-N2A2|Qe||dS2|,(2O)<strong>The</strong> current density J in each conductor is'and90Ji - -|pi-K- J2 - "K-l"-•Pu|-|pl-|-Nl|Oe| . |P2.|M N2IQ.I; Jl - -N, |Qe|v. ,J2- -N2|0e|u-.(21)(22)(23)<strong>The</strong>refore, the current in each conductor isI| =|Ji|A| - N,|0.|A,|v.| . l2=|j2|Aj - NzlOelAjlu.(24)Multiplication by the appropriate element <strong>of</strong> distance givesl||ds,|=N,A,|Oe||ds,||v.|.|Oe||v.|l2|dS2|=N2A2|0e||dS2(|u.|.|Qe||u.|.(25)<strong>The</strong>refore !-'dF,IIII '22 24m. c r-2(ds,ds2)*3(ds,ar)(ds2ar)(26)which is exactly the equation suggested by Ampere'5 in 1823.However.

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