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^{-39}x 3 x 10

^{ 8 }/ 5 x 10

^{6 }which works out to 2.481402 x 10

^{ –13 }eV. If we take 1 eV = 1.6 x 10

^{ –19 }J. Then in terms of Joules this would mean : 1.6 x 10

^{ –19}x 2.481402 x 10

^{ –13 }J = 3.969 x 10

^{ –32 }J. this is a phenomenally small amount of energy. To gain some idea of just how small this number is , if the positive of this number is taken it comes close to the number of atoms in the entire Universe ( 10

^{44}). Can this discrepancy really be ignored , because it means in effect that even if the field had 10

^{ 23}(i.e the number of free electrons in a conductor 10

^{ 23}cm

^{3}) photons in it the whole energy of the field would amount to hardly 10

^{ –9}J. While carrying out this calculation remember that 1 Coulomb ( or 1 Ampere ) of current means a flow of 6.25 x 10

^{ 18 }electrons /sec , so the figure of 10

^{ 23 }photons , going by the figures , could be representative of a current far in excess of 10

^{ 5}amps , as compared to what we actually have i.e , 3.969 x 10

^{ –32 }J. Yet this same field is supposed to give rise to currents that are 98% of the original current. That is the induced current in the secondary can be as much as 98% of that in the primary. To me it doesn’t make sense , especially because qualitatively there is supposed to exist no difference between the inductive field and the radiative field except for the distance represented by wavelength/2pi.