# Flux through an inclined area

Tags:
1. Jun 11, 2015

### niggchao

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

Φ = ∫B⋅dS

3. 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?

2. Jun 12, 2015

### rude man

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.

Last edited: Jun 12, 2015
3. Jun 12, 2015

### rude man

What relates polar (r, θ) to cartesian coordinates x and y? r is not a function of z.

Last edited: Jun 12, 2015
4. Jun 12, 2015

### niggchao

and http://www4f.wolframalpha.com/Calculate/MSP/MSP6082070f83052eaehe200002h13c1i6iciah25i?MSPStoreType=image/gif&s=40&w=93.&h=28. [Broken] then
B = (- sinφ + cosφ) μoI/2π*sqrt(x2+y2)
dS = (-[PLAIN]https://upload.wikimedia.org/math/8/7/5/875326710761d5ed42bdc6e30c4bf962.pngcos45 [Broken] - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) [Broken] 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?

Last edited by a moderator: May 7, 2017
5. Jun 12, 2015

### rude man

I think you're off to a great start!

B = (- sinφ + 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 [Broken] - [PLAIN]https://upload.wikimedia.org/math/8/b/a/8baf9dc7043aae61e37e171dc9f537e9.pngsin45) [Broken] 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.

Last edited by a moderator: May 7, 2017
6. Jun 12, 2015

### niggchao

Not really sure if I understood the dS part but I'll try

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

I did integrate dS and got 3√2 but I'm afraid this might be a coincidence. What do you think?

Last edited by a moderator: May 7, 2017
7. Jun 13, 2015

### rude man

which if you'll permit me = - k - i
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.

Last edited by a moderator: May 7, 2017
8. Jun 13, 2015

### niggchao

Thank you sir! This wolfram alpha is amazing. I didn't know something like this exists. Thanks again!

9. Jun 13, 2015

### rude man

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.