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Lorentz force calculation question.


DandelionTheory

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i need some help answering this question about calculating force on a portion of a coil.

got it from here: 

if lines DK < KC, lines AJ < JB, lines AJ = DK,and lines AD = BC

i can calculate the magnetic field from each side at point P with:

B = [μ0I/4π(L/2)] [Sinα +Sinβ] (the same α, L and β variables used in the video)

how do i calculate the force on coil C1 line JK if both coils magnetic fields come out of the screen? (picture, not video)

question.thumb.png.8eb50f6adf84fa261f5448f5e76c9c46.png

-DT

Edited by DandelionTheory
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I have a solution.

With respect to the picture i attached to this reply, you have to calculate the net force of coil C2 on coil C1.

if we assign points, with respect to the rules in the OP, we can assume coil section JK of C1 will experience a greater force from coil C2 section AD than coil C2 section BC will experience from coil C1 section JK.

 buuuuut you want math, so...

to calculate net force coil C2 induces on coil C1 you must brake up each side of coil C2 and calculate its force on each side of C1.

so the total net force of coil C2 on coil C1:

Ftotalnet = FnetAJ + FnetJK + FnetDK + FnetAD

now for a perpendicular current we need integrate, because the farther you move from a magnetic field source the weaker it becomes. to calculate the force on each section of coil C1 when the current section of C2 is perpendicular, we have to break up the length into arbitrary lengths d, calculate the force on each length, and add the net force.

 

FnetAJ = FnetAD + FnetBC FnetDC

         FnetAD dFAJ = I1dLAJ x 0I2/2πdr)

         FnetBC = dFAJ = I1dLAJ x 0I2/2πdr)

         FnetDC = I1b(μ0I2/2πrAJ)

FnetJK = FnetAB + FnetCD + FBC + FAD         

         FnetAB = dFJK = I1dLJK x 0I2/2πdr)

         FnetCD = dFJK = I1dLJK x (μ0I2/2πdr)

         FBC = I1b(μ0I2/2πrJK)

         FAD = I1b(μ0I2/2πrJK)

FnetDK = FnetAD + FnetBC + FAB

         FnetAD dFDK = I1dLDK x 0I2/2πdr)

         FnetBC = dFDK = I1dLDK x 0I2/2πdr)

         FAB = I1b(μ0I2/2πrDK)

FnetAD = FnetAB + FnetD+ FBC + FAD

         FnetAB dFAD = I1dLAD x 0I2/2πdr)

         FnetDdFAD = I1dLAD x 0I2/2πdr)

         FBC I1b(μ0I2/2πrAD)

         FAD I1b(μ0I2/2πrAD)

where:

I=current of coil C1(Amps)

I= current of coil C2(Amps)

rxx = distance between currents being calculated(meters)

b = length of coil section C2 working on C1(vector)

r = distance between current section being calculated and I2

 

now if you need to calculate the net force of coil C1 on coil C2, you would apply the same equations to each coil sides respectively.

But you'll find the net force acting on C1 is greater than the force acting on C2.

 

-DandelionTheory

with points.png

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i made a mistake

Lxx = length of the section of coil C1 being acted on.

also the integrals need to be edited, i didn't have the presence of mind at the time to see this mistake. i'm glad i have a friend to point it out. i have corrected the following variables because the second post cannot be edited any longer.:

         FnetAD dFAJ = I1dLAD x 0I2/2πdr)

         FnetBC = dFAJ = I1dLBC x 0I2/2πdr)

         FnetAB = dFJK = I1dLAB x 0I2/2πdr)

         FnetCD = dFJK = I1dLCD x (μ0I2/2πdr)

         FnetAD dFDK = I1dLAD x 0I2/2πdr)

         FnetBC = dFDK = I1dLBC x 0I2/2πdr)

         FnetAB dFAD = I1dLAB x 0I2/2πdr)

         FnetDdFAD = I1dLDC x 0I2/2πdr)

Edited by DandelionTheory
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