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Simplify my integrand to make the function Integrate[] work?

Posted 17 days ago
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To calculate a physical field, I need to integrate a complicated expression, with Sin and Cos and some 3/2 power. But when I run the Integrate function, it seems running forever. So what could I do to make it possible to get the Integrate result?

My code is simple:

magneticField = 
 Subscript[\[Mu], 0] * mI * 
  r * (Cross[{0, 0, 1}, {x - r*Cos[\[Theta]], y - r*Sin[\[Theta]], 
      0}])/EuclideanDistance[{x, y}, {r*Cos[\[Theta]], 
      r*Sin[\[Theta]]}]^3 // ComplexExpand // FullSimplify

Integrate[magneticField, {\[Theta], 0, 2*Pi}, 
 GenerateConditions -> False, 
 Assumptions -> (\[Theta] | x | y) \[Element] Reals]

enter image description here

2 Replies

I notice one issue: I think the setup isn't quite correct. Symmetry suggests that the B-field in the x-y plane should only be in the z-direction.

Given the Biot-Savart law, I get as my integrand

(\[Mu]0 mI)/(4 \[Pi])
   Cross[{-a Sin[\[Theta]], a Cos[\[Theta]], 0}, {x - a Cos[\[Theta]],
     y - a Sin[\[Theta]], 0}]/
  Norm[{x - a Cos[\[Theta]], y - a Sin[\[Theta]], 0}]^3 // Simplify

{0, 0, (a mI \[Mu]0 (a - x Cos[\[Theta]] - y Sin[\[Theta]]))/(
 4 \[Pi] (Abs[x - a Cos[\[Theta]]]^2 + Abs[y - a Sin[\[Theta]]]^2)^(
  3/2))}

Now, because of the symmetry of our situation, I assume without loss of generality that y=0. Now I do an indefinite integral (because for whatever reason the definite integral still seems to hang after this assumption), and I have

temp = Integrate[(a mI \[Mu]0 (a - x Cos[\[Theta]]))/(
  4 \[Pi] (a^2 + x^2 - 2 a x Cos[\[Theta]])^(3/2)), \[Theta]];
temp2 = (temp /. \[Theta] -> 2 \[Pi]) - (temp /. \[Theta] -> 0)

This returns an answer in terms of EllipticE and EllipticK. As a sanity check, I verify that the field at the origin is as expected,

In[5]:= temp2 /. x -> 0

Out[5]= (mI \[Mu]0)/(2 Sqrt[a^2])

and that the field displays the expected behavior as x goes to a, namely that it should be positive, diverge at the coil, and then go negative. Demonstration of code

Posted 16 days ago

Thank you, Duncan, I really love your answer. I learned a lot from your code and explanation. Thank you!

And sorry for my mistake in my expression. :P

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