What is the flux density at point p?

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SUMMARY

The flux density at point p due to a 3-A current in wire 2 is calculated using the formula B = I x 10^-7 (Integral sign) sinΘ/r^2 dl. The correct answer for the flux density is 2.45 x 10^-5 Wb/m^2, and the direction of the flux at point p is into the page. The discussion highlights the importance of correctly setting up the integral and considering the geometry of the problem, particularly the variable distance from the wire to the point of interest.

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Adventure123
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Homework Statement




1)
What is the flux density at point p in Fig "see link below". due only to the 3-A current in the portion of wire 2 shown?

1) 1.64 x 10^-5 Wb/m^2
2) 2.45 x 10^-5 Wb/m^2
3) 3.11 x 10^-5 Wb/m^2
4) 3.65 x 10^-5 Wb/m^2

2)
In which direction does the flux at point p of Question 1 point
1) Out of the page.
2) To the left of the page.
3) To the right of the page.
4) Into the page.

Fig Link
sellmontana.com/math/math.jpg
ignore the red marks.

Answers 2 and 2.. Is this correct?

Thanks

Homework Equations



B=I x 10^-7 (Integral sign) sinΘ/r^2 dl

The Attempt at a Solution



B= I x 10-7 (Integral sign) sin Θ /r^2 rdΘ/sin0

B= I x 10^-7/0.0200 (Integral sign) sinΘ dΘ

B=.5 x 10^-7 [-(-2)-(-5)] = ??

Does anyone have a real good document on how to Find flux density at a point? My current book does not fully explain it. Is there a online calculator that would help?

Thanks.
 
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Hi Adventure123,

I think you have a couple of problems with your integral. It's difficult for me to tell because I don't know how you set up the geometrey, but here are some notes:

In the first line of section 3, it looks like you are writing dl in terms of d(theta). That's good, but since your expression for dx has r in it, it looks like you forgot to take into account dr (since r is also variable). If so, it would probably be better to write dl in terms of theta and the perpendicular distance from the wire to the point (2 cm in your diagram). If we call that distance x, then dx would be zero.

In your second line of section 3, you brought r out of the integral, but it depends on theta and so cannot be removed like that. (If you write dl in terms of x, that constant x value would come out of the integral.)

But like I said, I could just be misinterpreting your method.
 

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