Ampere's Law problem help

In summary, we discussed the magnetic field created by a solid wire with current using Ampere's circuital law. By integrating the current density along the surface of the wire, we can find the magnetic field using the equation \oint_{c}H\cdot\,dl=\int_{s}J\cdot\,dS. The problem was not as difficult as initially thought.
  • #1
robert25pl
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Magnetic field due to a solid wire of current using Ampere's circuital law. Current flows with density [tex]J=J_{o}\frac{U}{r}a_{z}[/tex] [A/m^2] along long solid cylindrical wire of radius a having the z-axis as its axis. Find H

[tex]\oint_{c}H\cdot\,dl=\int_{s}J\cdot\,dS[/tex]

Am I on the right way to solution?

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

Thank You very much for any help?
 
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  • #2
I just finish this problem. It was not that difficult as I thought at the beginning. Thanks
 
  • #3


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}\
 

1. What is Ampere's Law and how does it relate to electromagnetism?

Ampere's Law is a fundamental law in electromagnetism that describes the relationship between the electric current and the magnetic field it creates. It states that the line integral of the magnetic field around a closed loop is equal to the current passing through that loop multiplied by a constant known as the permeability of free space.

2. What are some common problems that involve Ampere's Law?

Some common problems that involve Ampere's Law include calculating the magnetic field inside a solenoid or a long straight wire, determining the direction of the magnetic field around a current-carrying wire, and finding the force on a current-carrying wire in a magnetic field.

3. How do I apply Ampere's Law to solve a problem?

To apply Ampere's Law, you first need to identify the closed loop or path along which the line integral of the magnetic field will be calculated. Then, you need to determine the current passing through this loop and the direction of the magnetic field at every point along the loop. Once you have all the necessary information, you can use the equation B∫dl = μ₀I to solve for the magnetic field at a specific point.

4. What are the units for the constant μ₀ in Ampere's Law?

The constant μ₀, also known as the permeability of free space, has units of newtons per ampere squared (N/A²) or henrys per meter (H/m).

5. Are there any real-life applications of Ampere's Law?

Yes, Ampere's Law has many real-life applications. It is used in the design and operation of electric motors, generators, and transformers. It is also essential in understanding the behavior of magnetic materials and in the development of technologies such as magnetic levitation and magnetic resonance imaging (MRI).

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