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authorAnton Luka Šijanec <anton@sijanec.eu>2022-04-22 00:16:48 +0200
committerAnton Luka Šijanec <anton@sijanec.eu>2022-04-22 00:16:48 +0200
commit0333cfb34f7c91a700b69b71edf639604f5cfd0c (patch)
tree73ffcca17a86305c146a3b3dcacab44555419a41 /fiz/naloga/dokument.tex
parentnekaj dela fiz proj (diff)
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-rw-r--r--fiz/naloga/dokument.tex2
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}}