Sorry, didn't see any option to reduce the picture when uploading :/

Thanks for the comeback. Not too clear, would need a diagram.


However, confirmation of the poles shown in the video is if you place a magnet over the energized coil.
In this case I have drawn it inside but can be on any side of the coil... see the N pole locates adjacent
to the N pole of the coil. Now everyone knows like poles repel, right?


So the S pole that the hall sensor indicates on the top inside is present.


OK, I have redone the 'sketch' to better indicate all the block walls present in an air cored
rectangular coil. The external magnet shows the natural attachment direction.

poyntie's  cored solenoid coil is invalid in this discussion as my video is a representation of the polarity
present around a cross section of an air cored coil, which enhances Luc's original diagram, not the whole external field of a solenoid coil, which is a diversion.
 
Ron

Hey Iron

Below are pics to simulate 1 leg of the coil you used in the vid, showing only 6 wires for simplicity of making the diagram.  But still square like your coil.

These pics depict what you were showing. Just a cutaway of 1 leg of your coil.

More in next post

Mags

Now, whether it be N or S pole fields detected by the sensor, the lines of force are measured/detected  by going through the face of the sensor, the larger flat sides. If the field lines go through the edges, the sides of the sensor, top to bottom or side to side, there will be little to nothing detected or measured.

So below shows the lines of force around the coil leg with the sensor orientation as you have demonstrated. Notice how the lines of force do go through the face of the sensor when at the left or right, but in the middle, the lines go through the top to bottom of the sensor. Lets call the end of the sensor with the 3 leads the bottom of the sensor.

More next post

Mags

Ron, please do keep up your experiment and share your findings as it is worth more then words.

Khwartz, can you please calculate how much Joule energy it takes at Unity to lift 2.35Kg. 1mm, 2mm and 3mm in 1 second.

Thanks for your time and help

Luc

Yep! Except that if "by second" it means you ask for Watts and not Joules ^_^ but I will give you both:


□ W [J] = M [kg] × g [m.s^-2] × h [m]

□ P [W] = W [J] / T [ s]


● For 1 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 1/1000 [m]

= 23.5/1000 [kg.m^2.s^-2] = 0.0235 [J]


P [W] = 0.0235 [J] / 1 [ s] = 0.0235 [W]


● For 2 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 2/1000 [m]

= 47/1000 [kg.m^2.s^-2] = 0.047 [J]


P [W] = 0.047 [J] / 1 [ s] = 0.047 [W]


● For 3 mm height:

W [J] = 2.35 [kg] × ~10 [m.s^-2] × 3/1000 [m]

= 70.5/1000 [kg.m^2.s^-2] = 0.0705 [J]


P [W] = 0.0705 [J] / 1 [ s] = 0.0705 [W]


So the potential gravitational energy of the mass is increased proportionally to the height and the power in Watt goes with the energy in Joules, in term of value (but not the same nature of quantity).

If not clear enough, please just tell me

You're most welcome for the calculations 

The big picture is already deleted.

Nice that you are no more worried about your guiding system to stick the outside magnet between the your steel plates and that you have well advance on

Exactly,  if about "potential energy", the duration of the lifting doesn't matter, indeed. BUT if it was about "power" (Watts) it would be.

Nevertheless, it is not always completely true in practice: if we ask an electric motor to lift very fast a weight it will have non-proportional loses and the energy consumed will be a little bit more. But we are talking about energy CONSUMPTION

But for your simple lifting experiments I don't think it is relevant to care care if only electromagnets.

IMPORTANT NOTE:

The calculation W = M.g.h gives the change of potential energy in the gravitational field. It means that when the mass is on the floor its "gravitational potential energy" respect to the floor is zero, and when lift of any height it may restitue some energy when falling. This is this "potential falling energy" we have calculated.

Ron, please do keep up your experiment and share your findings as it is worth more then words.


Luc

Always a pleasure to work with you Luc.


I have touched up your graphic to show that you are quite correct. Although what you show is a simplification the outcome is the same as my crude sketches. So across the top of the graphic we
have a South pole, North pole (centre) South pole.


I have added the major Bloch walls in yellow. I have indicated the poles on the outer sleeve as if it was metal. (I know this was not the case) This ties in with the original E core graphic on page one where on DC we would have a S, N, S, on the E core legs, utilizing the the outer field of the coil.


What my sketches show (I hope) is how we can have the core and the coil both having the same N pole.
Well simply put, the 'hidden' south pole is generating the North pole.


So you are right on with your original graphic Luc, good work as always!!!


Ron