Proving Total Current through Insulating Wire Using Spherical Coordinates

In summary, the problem is to prove that the total current flowing through an insulating wire in the z direction, represented by J(p) = (I/pi) p^2 e^-p^2 in standard spherical polar coordinates, is equal to I. The attempt to use the idea J.dS = I and integrate with the scale factor taken into consideration did not yield the correct result. There is a missing factor of \rho^2 in the original equation, and the integral over the cross-sectional area of the wire may be written as I(\rho)=-I_0e^{-\rho^2}(\rho^2+1)+I_0, but it cannot be determined if this will result in the desired current of
  • #1
iontail
24
0
i am trying to solve this problem which states that

J(p) = (I/pi) p^2 e^-p^2 in z direction
is the current density flowing in the vicinity of insulating wire.
pi = pie

in standard spherical polar coordinates.

J is the current density.

I need to prove that the total current flowing through the wire is I.

I have tried to used the idea J.dS = I where

and integrate(i have taken the scale factor into consideration) but it does not yield the right result. Any suggestion on a way to move forward will be appreciated.
 
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  • #2
What are the limits of integration here? Still, I can't see how you can get a current of I because of your exponential, except if your relationship is incorrect.

[tex]J(\rho)=\frac{I}{\pi}e^{-\rho^2}[/tex]

If we have a wire of infinite size then this one would work, but this is just playing around.
 
  • #3
You missed a factor of [itex]\rho^2[/itex]:

[tex]J(\rho)=\frac{I}{\pi} \rho^2 e^{-\rho^2}[/tex]
 
  • #4
jtbell said:
You missed a factor of [itex]\rho^2[/itex]:

[tex]J(\rho)=\frac{I}{\pi} \rho^2 e^{-\rho^2}[/tex]

That's his original equation yes, whose integral over the cross-sectional area of the wire I think will be

[tex]I(\rho)=-I_0e^{-\rho^2}(\rho^2+1)+I_0[/tex]

Seeing as the OP has not given us the dimensions of the wire we can't go any further than that but I do not see how any normal choice of radius would allow the current to come out to be exactly I_0. I was just mentioning that if the \rho^2 dependence was dropped then, the integral would come out provided an infinite radius but as I stated I was not seriously suggesting that was an answer.

Another question is what does Stokes' Theorem have to do with the problem. I feel that there were some steps leading up to this point that the OP may have left out.
 
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1. What is the Stokes theorem?

The Stokes theorem, also known as the generalized Stokes theorem, is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface. It is an extension of the more well-known Green's theorem and is named after Irish mathematician George Gabriel Stokes.

2. How is the Stokes theorem used in physics?

The Stokes theorem is used in physics to calculate the flux of a vector field through a closed surface, which is a fundamental concept in electromagnetism. It is also used in fluid dynamics to calculate the circulation of a vector field around a closed curve. In general, the theorem is used to simplify and solve complicated integrals involving vector fields.

3. What is the relationship between Stokes theorem and current?

Stokes theorem is closely related to the concept of current in physics. In electromagnetism, the line integral of the electric field around a closed path is equal to the rate of change of the magnetic flux through the surface enclosed by the path. This is known as Ampere's law, which is a special case of the Stokes theorem. Therefore, Stokes theorem is often used to calculate the current in a circuit or the magnetic field produced by a current-carrying wire.

4. How does Stokes theorem relate to the conservation of energy?

Stokes theorem is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another. This principle is reflected in the theorem, as the surface integral and line integral in the theorem are equivalent, demonstrating that the same amount of energy is conserved in both forms.

5. Can Stokes theorem be applied to any type of surface or vector field?

Stokes theorem can be applied to any smooth surface and any smooth vector field, as long as the surface is closed and the vector field is continuous and has a continuous first derivative. However, the theorem may not hold for surfaces with sharp corners or discontinuous vector fields. In these cases, a modified version of the theorem, known as the generalized Stokes theorem, must be used to accurately calculate the integrals.

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