I find it a bit on the convoluted side, but... you can find the point by using paths instead of draws, do an inverse clip in a scope (adapted from here), re-draw the lines, exit the scope, and add the jump crossing:
\documentclass{beamer} \usepackage{tikz} \usetikzlibrary{intersections} \usetikzlibrary{calc} \usebackgroundtemplate{% \tikz\node[opacity=0.1] {\includegraphics[height=\paperheight,width=\paperwidth]{example-grid-100x100pt}};} \usepackage{circuitikz} \newlength{\crossing} \makeatletter \setlength{\crossing}{\ctikzvalof{bipoles/crossing/size}\pgf@circ@Rlen} \makeatother \tikzset{ clip even odd rule/.code={\pgfseteorule}, % Credit to Andrew Stacey invclip/.style={ clip,insert path= [clip even odd rule]{ [reset cm](-\maxdimen,-\maxdimen)rectangle(\maxdimen,\maxdimen) } } } \begin{document} \begin{frame} \begin{tikzpicture}[scale=4,transform shape] \path[name path=p1] (0,0) -- (2,0) ; \path[name path=p2] (1,1) -- (1,-1) ; \begin{scope}[overlay] \path [name intersections={of=p1 and p2,by=cross}]; \clip [invclip] (cross) circle[radius=0.5\crossing]; \draw (0,0) -- (2,0) ; \draw (1,1) -- (1,-1) ; \end{scope} \path (cross) node[jump crossing,rotate=-90]{}; \end{tikzpicture} \end{frame} \end{document} 
(okular is struggling with the antialiasing here a bit...)