How Does Ampere's Law Apply to a Solid Wire with Non-Uniform Current Density?

[robert25pl]

Magnetic field due to a solid wire of current using Ampere's circuital law. Current flows with density J=J_{o}\frac{U}{r}a_{z} [A/m^2] along long solid cylindrical wire of radius a having the z-axis as its axis. Find H

\oint_{c}H\cdot\,dl=\int_{s}J\cdot\,dS

Am I on the right way to solution?

\int_{s}J\cdot\,dS =
=\int_{r=0}^{r} \int_{\Phi}^{2\pi}J_{o}\frac{U}{r}a_{z}\cdot\ r \ dr\, d\phi\,a_{z}

Thank You very much for any help?

 

[robert25pl]

I just finish this problem. It was not that difficult as I thought at the beginning. Thanks

 

[thepatientmental]

Yes, you are on the right track! Let's break down the steps to solve this problem using Ampere's law.

Step 1: Identify the given information and parameters
In this problem, we are given the current density J, which is a function of the radius r and the constant J_o. The current is flowing along a long cylindrical wire with radius a and the z-axis as its axis. We need to find the magnetic field H at a distance r from the wire.

Step 2: Apply Ampere's law
Ampere's law states that the line integral of the magnetic field H along a closed path is equal to the total current passing through the surface bounded by that path. In this case, the path is a circle of radius r, centered at the wire. So, we can write the equation as:

\oint_{c}H\cdot\,dl=\int_{s}J\cdot\,dS

Step 3: Simplify the equation
We can simplify the equation by using the symmetry of the problem. Since the current is flowing along the z-axis, the magnetic field will also have a component only along the z-axis. So, we can write:

\oint_{c}H\cdot\,dl=\int_{s}H_{z}\cdot\,dl

Also, the surface integral can be simplified as:

\int_{s}J\cdot\,dS=\int_{s}J\cdot\,dS\,a_{z}

Step 4: Solve the integral
Now, we can solve the integral by substituting the given values and integrating over the surface. We get:

\int_{r=0}^{r} \int_{\Phi}^{2\pi}J_{o}\frac{U}{r}a_{z}\cdot\ r \ dr\, d\phi\,a_{z} = J_{o}\frac{U}{r} \int_{0}^{2\pi} \int_{0}^{r} r \ dr\, d\phi\,a_{z} = J_{o}\frac{U}{r} \pi r^2 a_{z}

Step 5: Solve for H
Now, we can equate the two equations and solve for H. We get:

\oint_{c}H\cdot\,dl=\int_{s}H_{z}\