How Does Coil Orientation Affect Mutual Inductance Calculation?

AI Thread Summary
The discussion focuses on calculating the mutual inductance between a rectangular coil and an infinite straight filament. Participants explore the conversion of magnetic field and area elements into Cartesian coordinates to facilitate integration. There is confusion about expressing the distance variable 'r' in terms of x, y, and z, and how to correctly define the area element dS. Suggestions include using Wolfram Alpha for complex integrals, and participants express appreciation for its utility in solving mathematical problems. The conversation emphasizes the importance of careful integration and the potential messiness of closed-form results.
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Homework Statement


A rectangular coil is composed of 150 turns of a filamentary conductor. Find the mutual inductance in
free space between this coil and an infinite straight filament on the z axis if the four corners of the coil
are located at
2gud0eg.png


Homework Equations



B = (phi-hat) μoI/2πr
dS = (n-hat) dydz

2dgkvv5.png
[/B]

Φ = ∫B⋅dS

The Attempt at a Solution


I can easily solve this problem if the coil is parallel to the y-z plane and x = 1 since the r in the formula varies only on y thus, becomes sqrt(y2 + 1) then convert phi-hat to cartesian unit vectors and the area unit vector is just -(x-hat).

Now, in this problem, I have converted the area unit vector to its cartesian unit vector. I am confused on how I will write the expression for r since it now varies on x, y, and z?
 
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I would suggest the following:
Express the B field and the element of area dA in cartesian coordinates. Then form the integral φ = ∫B⋅dA and you know the rest I assume.
The element dA has a vector normal to the described area & will consist of an i and a k component as I think you can see. The B field will have an i and a j component.
i, j and k are unit vectors in the x, y and z directions. Vectors are in bold.

There may be a shortcut method but I don't see one at the moment.
 
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niggchao said:
I am confused on how I will write the expression for r since it now varies on x, y, and z?
What relates polar (r, θ) to cartesian coordinates x and y? r is not a function of z.
 
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79eb09cbb46123591264cf57618a757b.png
and http://www4f.wolframalpha.com/Calculate/MSP/MSP6082070f83052eaehe200002h13c1i6iciah25i?MSPStoreType=image/gif&s=40&w=93.&h=28. then
B = (-
8baf9dc7043aae61e37e171dc9f537e9.png
sinφ +
7b5fdad25716879ce0706bf95efd82c0.png
cosφ) μoI/2π*sqrt(x2+y2)
dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) dydx I changed it from dydz to dydx because my variables in the integral are x and y. Is this okay?

Then integrate the dot product to get the flux.
What do you think?
 
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niggchao said:
79eb09cbb46123591264cf57618a757b.png
and http://www4f.wolframalpha.com/Calculate/MSP/MSP6082070f83052eaehe200002h13c1i6iciah25i?MSPStoreType=image/gif&s=40&w=93.&h=28. then
B = (-
8baf9dc7043aae61e37e171dc9f537e9.png
sinφ +
7b5fdad25716879ce0706bf95efd82c0.png
cosφ) μoI/2π*sqrt(x2+y2)
dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) dydx I changed it from dydz to dydx because my variables in the integral are x and y. Is this okay?

Then integrate the dot product to get the flux.
What do you think?
I think you're off to a great start!

B = (-
8baf9dc7043aae61e37e171dc9f537e9.png
sinφ +
7b5fdad25716879ce0706bf95efd82c0.png
cosφ) μoI/2π*sqrt(x2+y2),
what are cosφ and sinφ in terms of x and y?

dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) dydx is almost right. This is the tricky part. Your area element is dy times the differential distance along the sloping area. Call it dζ. So your differential area is dydζ.
Can you see that dζ2 = dx2 + dz2 ? But you can eliminate dz; what is it in terms of dx? Your differential area can then be (constant)*dxdy, multiplied by the normalized normal vector S. To check that you got the right constant, just integrate (constant)*∫∫dy dx over the limits given you. You know the area has to come out to 3√2.

The rest is as you say. Don't forget N = 150 and mutual inductance is a positive number.
 
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Not really sure if I understood the dS part but I'll try

wi57ro.png

cos45 = dx/dζ
dζ = dx/cos45
dS = (1/cos45) dxdy

dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) (1/cos45) dxdy

I did integrate dS and got 3√2 but I'm afraid this might be a coincidence. What do you think?
 
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niggchao said:
dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) (1/cos45) dxdy
which if you'll permit me = - k - i :smile:
I did integrate dS and got 3√2 but I'm afraid this might be a coincidence. What do you think?
I think it's no coincidence. I think you're banging at the door.
So now, proceed to do the integration? A word of warning: there are of course two ways to do the double integration over x and y. You will be appalled at how messy the closed-form results are either way. BUT - if you do what I did and let wolfram alpha do the definite integrals you happily get the same numerical answer either way, as of course you should. So if you get an answer we can compare notes.
 
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Thank you sir! This wolfram alpha is amazing. I didn't know something like this exists. Thanks again!
 
niggchao said:
Thank you sir! This wolfram alpha is amazing. I didn't know something like this exists. Thanks again!
Yes, WA is a miracle. I hope you have access to the Pro version. As a retiree I can't afford the premiums but luckily I didn't need too much computing time for the integrals. Nice work.
 
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