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I am trying to convert $\frac{dB\mu A}{m}$ to $\frac{dB\mu V}{m}$

So, I know that;

$\frac{dB\mu V}{m}$ = $\frac{dB\mu A}{m} + 51.5$

However, I cannot find a source explains where 51.5 comes from. Is it related with air impedence? Thank you for explanations in advance.

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  • $\begingroup$ What are $d$, $B$, $\mu$, $A$, and $m$? $\endgroup$ – garyp May 26 at 13:23
  • $\begingroup$ dB = decibel $\mu$ = micro A = ampere V = Volt $\endgroup$ – Zapdos May 26 at 13:25
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It comes from the freespace impedance, convert it to same scale (20*log10(377Ω)) = 51.5

In dB multiplication becomes adition, realizing this, you have something that looks like ohms law.

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