DandelionTheory

What happens when you apply force on 2 sides of a triangle? Some of it is applied to it's center axis and cancels. Did you forget the source of B is larger than I? I talked about fringe fields and you attempted to proof one axis when I said you had some canceling on the sides. How many sides are there? Do you think 2 evens and an odd make an odd? Where's your math that shows F2? Did you do the vector calculation or are you assuming with "new physics"? Damn now I gotta be the only one that cares. Science died when you all stopped dreaming.

I need some help with the calculus involved with current at right angles from itself. The picture shows a basic scenario where the calculation would apply, and a wire setup (fig5) to attempt to convey the next step. -DandelionTheory

I appreciate your help and patience. 😊 I expected fields to interact, I do not have the ability to explain the balanced forces of I1, I2 & I3 on B4, B5 & B6. I only suspect they do due to some fancy right hand rule gestures. I was going with the fact that steel beams are used as passive shielding in MRI machines. http://mriquestions.com/how-to-reduce-fringe.html

right, I1 will push on B4 and B5 inwards, I2 and I3 will push on B4, B5 and B6 outward. i wonder if they.... cancel out, but i cant know that till its calculated. I need help with that part. weather its fringe fields or B4, B5 & B6 acting on each other

i would like to see how you concluded your internal force calculation. in one of the illustrations i attempted to "hide" like magnetic fields within the steel core, the bottom current interacts with the inside of the iron core through "fringing fields" which do not have a permeability of iron because its an air core.

yes. what else would i need to calculate? i have other versions of lifters.

Ive added sources to the magnetic fields with this new picture.

Equilateral or isosceles with the odd current at the odd angle vertex. The E vectors have no X angle. I made a correction.

My tutor told me I need to understand more calculus before derivatives can be understood. They also explained my Z angles were off for my B vectors, and I needed to specify my vector notation. So arbitrary vector V = V<magnitude, angle from X origin, angel from Z origin> I still don't know how to calculate induced magnetic field in steel like the ?? State in the second picture below. If I can answer that, maybe I can answer how the initial B fields of the first picture are generated 1st 2nd -DandelionTheory

I set up a g vector at the top. Breh read all the variables please. Gravity is constant and in the y<270°,0°> direction. The original is still viable, it is still here and I am working on it still with you. The magnetic fields experience torque, I don't know how much and what exact direction yet till my tutor helps me or one of you are kind enough to help me solve for B3 in the second lifter.

I do not know how to calculate magnetic field due to current yet. The new lifter is to illustrate how the currents can be "attached" to the original lifter and be inline with the initial magnetic fields.

I drew another example I need to proof. Its just an idea, but I need some help figuring out how to calculate B3 if Q is steel.

I cannot answer that at the moment. I'm currently being tutored to express magnetic fields induced in iron due to a current loop.

Conservation of momentum is normally calculated as loose particles and their magnetic fields acting on the other particle. If a magnetic field independent of I is at the correct angle, acceleration in the B angle plus or minus 90° is "felt" by the current. I have not brought up how the uniform B fields are made initially, that's not the point of the problem. I have given values to magnitudes and some angles to show what I mean at the bottom right of the picture. PS, I'm currently unemployed. Lol. *Bong toke* -DandelionTheory

Each current has a force calculation based on the total magnetic field at the intersection of xy plane. if they are the same structure, total force is applied to the whole structure in one direction if one current is 180° from the other 2 on the Z plane.

I'm sorry, this one. I added mass and gravity to show the definition of "lifter"

Thank you. I will take the time to learn Latex and come back with editable equations. Until then, I have respecified the defined magnitude of B4 , B5 and B6 to be equal to the length of each respective parallel side of ∆ABD. The total force seems to always be within 180° of the line between the 2 currents with the same Z angles.

Yes. Do I need to specify the magnitude of each arbitrary B to be proportional to the sides of ∆ABD?

I figure "magnetic lifter" was appropriate name for this problem. The question is: "What is the total Lorentz force on all currents? Is my math correct for total force on all currents?

Are you considering charged fluid? This idea would mean any magnetic field would add to the change in momentum of the fluid, meaning all moving charge carrier's magnetic field would do work on any perpendicular charge carrier in the field.

it appears i have found something, but im still not too sure ive applied it correctly. in attempting to calculate the B field interacting with the current, i found a resource in calculating B field between the air gaps of an iron core transformer: https://en.wikipedia.org/wiki/Electromagnet in section "Magnetic field created by a current" it explains how to calculate the B field of an iron core magnet with an air gap. i would then use the Lorenz force to calculate the force experienced by the current. from what i can tell, the placement of a current next to its own dipoles is a weird concept. I'm assuming this produces a momentum change in the positive Y direction, but i can be wrong. ps, im looking for a tutor in math.

The magnetic force on a moving charge reveals the sign of the charge carriers in a conductor. A current flowing from right to left in a conductor can be the result of positive charge carriers moving from right to left or negative charges moving from left to right, or some combination of each. When a conductor is placed in a B field perpendicular to the current, the magnetic force on both types of charge carriers is in the same direction. from: https://www.britannica.com/science/Lorentz-force so if an aqueous salt has both charge carriers and the electric field applied was ac, the Lorentz force would mean each charge carrier interacts with another's magnetic field at a right angle. My point is the average in momentum does not balance out, and there is a net force inline with the right angle bisector.

so if copper sulfate was in a tube, where it had a 90 degree bend and electrodes were on each end of the tube electrically connected to the copper sulfate; i want to know if the moving charges are the same calculation when i attempt to do a summation of forces over time. because if the electric field is AC, the moving charges would always interact at the 90 degree bend. right? o.O

im attempting to understand what i can and cannot do; why i can and cannot do them, and what kind of responses i get by asking them. because i know its not normal, im trying to see if its correct. because my next question is can i move the dipoles to another place? and then how do i calculate the sum of forces over time on all structures in the system of figure 4?

i have drawn another picture to ask a specific question visually because of the new shape of the core. in the new picture (labeled as fig 2), i show the iron core as a capital I shape. are the indicated poles correct? what im doing is reshaping the magnet to reorient the dipoles. ill show you in the next one if this looks right, i know its tedious but you will have a better chance with fig 4.