moffy:
What makes you think the steel (core) isnt going to do much good?
Because the air gap is large. You don't have the field going across a small air gap to more steel which is connected back to the coil. But in asking this question you show you don't have a basic knowledge of inductors. You really need an inductor/electromagnet specialist as there are lots of details that need to be correct. The construction is quite specialised.
The permeability of soft iron may be as 50000 times that of a vacuum (ref https://en.wikipedia.org/wiki/Magnetic_core).What makes you think the steel (core) isnt going to do much good?
It is an inductor. It generates a magnetic field in response to a current. Energy stored is L*I*I/2. It will have some series resistance due to the windings, but primarily inductance. V = L*di/dt = N*dPHI/dt. So inductance and flux are directly related. Flux is your measure of guass.
Based on the equation, the permeability of the core is not as important (so long as it's large) as the proportion of the path that contains the core.But as long as you want to relate this to an inductor, you cant tell me that if you take an air core coil and measure the inductance with an without a 5000u core that the inductance isnt going to go up significantly, even if it is a straight coil that does not curve back to meet the other end.
Not particularly... I was just ignoring the fact that the path length of the air gap changes as the air is replaced by ironAlso, you two must still be looking at the properties of a finished inductor.
I just used the formula from above of [LATEX]B=\frac{NI\mu_0}{L_{gap}}[/LATEX] rearranged to [LATEX]N=\frac{BL_{gap}}{I\mu_0}[/LATEX]. B is in Teslas, and is therefore 0.5T for 5000 Gauss. I is 10A. Lgap was set to 6mm... but should actually be closer to 6cm, so perhaps you'll need 15km of wire instead.. Ok, that's enough mistakes for me for one night.120 turns at 10 amps gives 5000 Gauss...how did you calculate that?
Sounds good to me. You can draw the field lines from N to S with both the short magnet and the long magnet (i.e. the one with the iron) - the short one should have a high density of flux lines on the short sides, without much leaking out into the rest of the world while the long magnet will not have the high density and will therefore have the lines extended out further away from the magnet ends... That's my take on it anyway. I had to write an electrostatic solver for some numerical programming course in uni, which could be used to show the fields here, but I misplaced the code.One possible answer: the laminations provided a lower reluctance to the magnetic path between the magnet and the pin because of their higher permeability than air
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