The design you mention has no core. And even if an iron piece is inserted to fill the tube, the flux will have to return through the air. A long path-length through air means you will need thousands of times more amp-turns to achieve the same flux.
Summarizing the above: Flux = Amp-turns / Reluctance
where Reluctance ~ Mu0 * Area / (air-path-length).
If you can get your hands on donut shaped piece of core steel with one piece cut out of the donut (similar to horseshoe, but bent a little more), then you can insert the piece into the opening in the horseshoe.
You will have to monitor the current and temperature as you go to avoid exceeding insulation temperatures, starting a fire, or tripping a breaker. Current should vary in proportion to Voltage / [number-of-turns(^2)] for given geometry. Also highest current occurs when there is no piece inserted into the flux (air-path-lenght of flux is longest).
Just some thoughts and not intended as a substitute for adult supervision ;-)