Magnetic energy of a conductor with a hole

• crick
In summary, when calculating the magnetic energy of a cylindrical conductor with a hole, it is equivalent to calculating the energy of two separate conductors, one with a uniform current and the other with a hole, and then combining the energies. This is due to the superposition principle, which allows for the calculation of fields to be summed.
crick

Homework Statement

A cylindrical conductor of radius ##a## has a hole of radius ##b## and there is a current ##i## uniformly spread inside the conductor.

Calculate the magnetic energy of the conductor.

Homework Equations

$$U_{m}=\int_{Volume} \frac{B^2}{2 \mu_0} dV$$

The Attempt at a Solution

I know that this system is equivalent to a integer cylindrical conductor with uniformly spread current ##i_{+}##, plus another conductor in the place of the hole, with uniformly spread current ##i_{-}##, where

$$i_{+}=\frac{i \, \pi a^2}{\pi (a^2-b^2)} \,\,\,\, \mathrm{and} \,\,\,\, i_{-}\frac{i \, \pi b^2}{\pi (a^2-b^2)}$$

I see that on my textbook this exercise is solved by saying that the total energy would be the sum of the energy due to ##i_{+}## and the energy due to ##i_{-}##, that is (doing the integral above):

$$U_{m}=\frac{\mu_0}{16 \pi} (i_{+}^2-i_{-}^2)$$

But I do not agree with that. The superposition principle is valid in the calculation of the magnetic field but not in the calculation of the total energy.

Roughly said, the sum of squares is not the square of the sum, so shouldn't I instead firstly calculate the total magnetic field il all the conductor (summing vectorially the magnetic field due to ##i_{+}## and the one due to ##i_{-}## in every point) and then calculate

$$U_{m}=\int \int \int_{Volume} \frac{B_{total}(x,y,z)^2}{2 \mu_0} dx dy dz$$

?

This process looks more correct because the superposition principle is allowed while calculating fields.

Did I get this right is the process done on my textbook valid to calculate the total energy?

Attachments

• fggggggg.jpg
2.5 KB · Views: 416
• gggggggggggggggg.gif
11 KB · Views: 495
Do you get a different answer?

If two systems are physically equivalent, then they will follow equivalent rules.

1. What is magnetic energy?

Magnetic energy is a form of energy that results from the interactions between magnetic fields and electric currents. It is the potential energy stored in a magnetic field.

2. How does a conductor with a hole affect magnetic energy?

A conductor with a hole can alter the distribution of magnetic fields, thereby changing the amount of magnetic energy present. This is because the presence of the hole creates a change in the flow of electric current, which in turn affects the magnetic field.

3. Is the magnetic energy of a conductor with a hole different from a solid conductor?

Yes, the magnetic energy of a conductor with a hole will be different from a solid conductor due to the presence of the hole. The hole creates a change in the flow of electric current and thus alters the distribution of the magnetic fields and the amount of magnetic energy present.

4. How can the magnetic energy of a conductor with a hole be calculated?

The magnetic energy of a conductor with a hole can be calculated by using the formula E = 1/2 * μ * I^2 * A, where E is the magnetic energy, μ is the permeability of the medium, I is the current flowing through the conductor, and A is the area of the conductor.

5. What are some real-life applications of magnetic energy in conductors with holes?

Magnetic energy in conductors with holes is used in many everyday devices such as speakers, motors, and generators. It is also utilized in magnetic resonance imaging (MRI) machines in the medical field. Additionally, the concept of magnetic energy in conductors with holes is important in understanding the behavior of magnetic fields in various electrical systems and devices.

• Introductory Physics Homework Help
Replies
10
Views
352
• Introductory Physics Homework Help
Replies
11
Views
280
• Introductory Physics Homework Help
Replies
2
Views
247
• Introductory Physics Homework Help
Replies
2
Views
428
• Introductory Physics Homework Help
Replies
25
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
376
• Introductory Physics Homework Help
Replies
23
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
470
• Introductory Physics Homework Help
Replies
6
Views
252