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author | Anton Luka Šijanec <anton@sijanec.eu> | 2022-04-22 00:16:48 +0200 |
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committer | Anton Luka Šijanec <anton@sijanec.eu> | 2022-04-22 00:16:48 +0200 |
commit | 0333cfb34f7c91a700b69b71edf639604f5cfd0c (patch) | |
tree | 73ffcca17a86305c146a3b3dcacab44555419a41 /fiz/naloga/dokument.tex | |
parent | nekaj dela fiz proj (diff) | |
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Diffstat (limited to '')
-rw-r--r-- | fiz/naloga/dokument.tex | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/fiz/naloga/dokument.tex b/fiz/naloga/dokument.tex index c8cff2e..f3ba60e 100644 --- a/fiz/naloga/dokument.tex +++ b/fiz/naloga/dokument.tex @@ -105,7 +105,7 @@ Da dobimo enačbo za polje enega navitja Helmholtzove tuljave, enačbo za eno za $$B_1(z)=\frac{n\mu_0R^2I}{{2\left(z^2+R^2\right)}^{3/2}}\text{.}$$ Zanima nas taka vrednost $z$, ki je v središču med obema navitjema. To je, kot je zgoraj opisano, $z=R/2$: $$B_1\left(\frac{R}{2}\right)=\frac{n\mu_0R^2I}{2\left(\left(\frac{R}{2}\right)^2+R^2\right)^{3/2}}\text{.}$$ -Jakost magnetnega polja med obema simetričnima navitjima je dvakratniku $B_1$. +Jakost magnetnega polja med obema simetričnima navitjema je enaka dvakratniku $B_1$. $$B\left(\frac{R}{2}\right)=2{B_1}_Z\left(\frac{R}{2}\right)=$$ $$=\frac{\cancel{2}n\mu_0R^2I}{\cancel{2}\left(\left(\frac{R}{2}\right)^2+R^2\right)^{3/2}} =\frac{n\mu_0R^2I}{\left(\frac{1}{4}R^2+R^2\right)^{3/2}=\left(\frac{5}{4}R^2\right)^{3/2}} |