Surface integral for line current

In summary, the question asks to calculate the integral of a closed curve formed by the intersection of two surfaces, using a given field. The answer is obtained by converting to cylindrical coordinates and using Stokes' theorem, and the result is 2πF0a(8(√13−2)3/2+1). A mistake was made in the first approach by not accounting for scale factors in curvilinear coordinates.
  • #1
Incand
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Homework Statement


Calculate the integral ##\oint_C \vec F \cdot d\vec S##, where ##C## is the closed curve constructed by the intersection of the surfaces ##z = \frac{x^2+y^2}{4a}## and ##x^2+y^2+z^2=9a^2##, and ##\vec F## is the field ##\vec F = F_0\left( \frac{a}{\rho}+\frac{\rho^2}{a^2} \right)\hat \phi##. (##a## and ##F_0## are constants.)

(I'm guessing this fits better here than in the calculus forum since since the question is as preparation for EM-class)

Homework Equations


Line current with strength ##J##
##\vec F = \frac{J}{2\pi \rho} \hat \phi##.

The Attempt at a Solution


We see the first part of the force is a line current with strength ##2\pi F_0a## so it's total contribution to the integral over a closed loop is ##2\pi F_0a##.

I'm left with ##\vec F_2 = F_0\frac{\rho^2}{a^2}\hat \phi##. We note from ##z=\frac{x^2+y^2}{4a}## that if ##a >0## ##z## has to be strictly positive and if ##a < 0## ##z## is strictly negative.
Substituting into the second equation we have
##(z+2a)^2-4a^2 = 9a^2 \Longrightarrow z=(\pm \sqrt{13} -2)a##. We conclude that ##z## is constant and since ##\rho^2 = x^2+y^2##, ##\rho## is also constant.
Calculating the curl we get ##\nabla \times \vec F_2 = \frac{3F_0}{a^2}\rho \hat z##.
Using stokes theorem and since we know ##\rho## is constant and using ##\hat z## as the normal vector
##\int_C \vec F_2 \cdot d\vec r = \frac{3F_0}{a^2}\rho \int_S dS = \frac{3\pi F_0}{a^2}\rho^3 ##
Let's assume that ##a>0## we then have that ##\rho = 2a\sqrt{\sqrt{13}-2}##. Using this we have
##\int_C \vec F_2\cdot \vec r = 24\pi F_0a\left( \sqrt{13}-2 \right)^{3/2}##.
So the total value of integral should be
##2\pi F_0a\left(12(\sqrt{13}-2)^{3/2}+1\right)##.Another approach would be to use the line element ##d\vec r = d\rho \hat \rho + \rho d\phi \hat \phi + dz \hat z##
##\int_C \vec F_2 \cdot d\vec r = \int_C \frac{F_0\rho^3}{a^2}d\phi## And everything inside should be a constant. Since both ##z## and ##\rho## are constant we should be on a circle that's parallel to the ##xy##-plane and we should integrate from ##\phi =0 \to \phi = 2\pi##. So the integral becomes
##\int_C \vec F_2 \cdot d\vec r = \frac{2\pi F_0 \rho^3}{a^2} = 8\pi aF_0(\sqrt{13}-2)^{3/2}##. Giving us the value
##2\pi F_0a\left(8(\sqrt{13}-2)^{3/2}+1\right)##.

According to the answer the second approach gave me the right answer and the first one is wrong but I can't see where I go wrong?
 
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  • #2
Have you drawn a picture of what the contour looks like?
This is the intersection of a cone and a sphere.
Since your integral uses spherical coordinates, convert the contour to spherical coordinates as well. That should help.
 
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  • #3
RUber said:
Have you drawn a picture of what the contour looks like?
This is the intersection of a cone and a sphere.
Since your integral uses spherical coordinates, convert the contour to spherical coordinates as well. That should help.

Sorry I should've clarified better. The field is given in cylindrical coordinates (the book uses ##\rho## for cylindrical.)
Should be a paraboloid not a cone since we have ##z=\frac{x^2+y^2}{4a}## not ##z^2=\frac{x^2+y^2}{4a}## so the intersection should be a circle. I did make a drawing but I'm not too great at those. Do you think the second solution is correct or did I just happen upon the correct answer there? Be cause I feel I did most of the work similar for both approaches.
 
  • #4
Yes, you are z is growing with the square of the radius of the circle, not linearly with the radius like in a cone.
Your contour is a circle parallel to the xy plane. So the second method leaves fewer opportunities to make a mistake.
 
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  • #5
I realized what my error was here. When I calculated the surface integral i got from stokes theorem I had
##3F_0a^2\int_S \rho d\vec S##. S is a ##z##-surface so the surface element is given by ##d\vec S = \hat z \rho d\rho r\phi##. Using this we have (let ##\rho_0## be the radius of the circle as to not confuse them)
##3F_0a^2 \int_0^{\rho_0}\int_0^2\pi \rho^2 d\phi d\rho = 2\pi F_0\rho_0^3## yielding the correct answer in the first case as well. So the mistake I made was that I didn't accountfor that I have scale factors in curvilinear coordinates.
 

1. What is a surface integral for line current?

A surface integral for line current is a mathematical calculation used in electromagnetism to determine the magnetic field produced by a current-carrying wire over a given surface. It takes into account the direction, magnitude, and length of the current, as well as the orientation and size of the surface.

2. How is a surface integral for line current different from a regular surface integral?

A surface integral for line current is specifically used for calculating the magnetic field produced by a current-carrying wire, while a regular surface integral can be used for a variety of calculations in mathematics and physics. Additionally, the surface integral for line current involves a vector field, while a regular surface integral involves a scalar field.

3. What is the formula for calculating a surface integral for line current?

The formula for calculating a surface integral for line current is ∫∫(I x dL)/r², where I is the current, dL is the differential length element along the current, and r is the distance from the current to the surface element. This formula is also known as the Biot-Savart law.

4. What are some real-world applications of surface integral for line current?

Surface integrals for line current are used in a variety of real-world applications, such as designing and analyzing electrical circuits, calculating the magnetic fields of motors and generators, and understanding how current flows through wires and conductors. They are also used in medical imaging techniques, such as MRI machines, which use strong magnetic fields to create images of the body.

5. Can the surface integral for line current be used for non-linear current distributions?

Yes, the surface integral for line current can be used for both linear and non-linear current distributions. However, the calculation becomes more complex for non-linear distributions, as it involves integrating over the entire surface element instead of just the differential length element along the current. In these cases, numerical methods or advanced mathematical techniques may be necessary to calculate the surface integral accurately.

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