Energy density(Magnetic and Electric)

In summary, the energy density at the surface of the wire for the magnetic field is μI^2/8π^2r^2 and for the electric field is εI^2/2π^2r^4. These values take into account the effects of the surrounding medium on the magnetic and electric fields at the surface of the wire.
  • #1
zakaqel
10
0

Homework Statement




A length of Cu wire carries a current of 10 A uniformly distributed throughout its cross section. Calculate the energy density of (i) magnetic field (ii) the electric field at the surface of the wire. The wire diameter is 2.5 mm and its resistance per unit length is 3.3K/ km


Homework Equations


I know that the energy density of magnetic field:
U=uoI^2/16pi

I know that the energy density of electric field:
U=1/2*epsilon0I^2*R^2



The Attempt at a Solution



Using the above equations I found that the magnetic energy density of the wire is 2.5 X10^-6 J/m^3 and the electric energy density of the wire is 48.2 J/m^3.

How am I supposed to find the surface energy density(for both magnetic and electric) using the above values. Does it have anything to do with the radius?
 
Physics news on Phys.org
  • #2


I would like to offer some guidance on how to approach this problem.

Firstly, it is important to note that the energy density of a magnetic field is given by the equation U = (1/2)B^2/μ, where B is the magnetic field strength and μ is the permeability of the medium. Similarly, the energy density of an electric field is given by U = (1/2)εE^2, where E is the electric field strength and ε is the permittivity of the medium.

To find the energy density at the surface of the wire, we need to consider the fact that the current is uniformly distributed throughout the cross section of the wire. This means that the magnetic field and electric field will also be uniformly distributed throughout the wire. However, at the surface of the wire, the magnetic field and electric field will be affected by the surrounding medium (air, for example).

To take this into account, we can use the concept of boundary conditions. At the surface of the wire, the magnetic field and electric field must satisfy the boundary conditions of continuity and tangentiality. This means that the magnetic field and electric field at the surface must be equal to the magnetic field and electric field inside the wire, but taking into account the different permeability and permittivity of air.

To find the magnetic field at the surface, we can use the equation B = μH, where H is the magnetic field intensity. Since the current is uniformly distributed, the magnetic field intensity will also be uniformly distributed. Therefore, we can use the equation H = I/2πr, where r is the radius of the wire. Plugging this into the equation for the magnetic energy density, we get U = (1/2)(μI/2πr)^2/μ = μI^2/8π^2r^2.

Similarly, to find the electric field at the surface, we can use the equation E = ρJ, where ρ is the resistivity of the wire and J is the current density. Since the current is uniformly distributed, the current density will also be uniformly distributed. Therefore, we can use the equation J = I/πr^2. Plugging this into the equation for the electric energy density, we get U = (1/2)ε(J/πr)^2 = εI^2/2π^2r^4.

Using the
 
  • #3




To find the surface energy density for both magnetic and electric fields, you will need to use the surface area of the wire rather than the volume. This is because the energy density is a measure of energy per unit volume, so we need to consider the energy at the surface of the wire rather than throughout its entire volume.

For the magnetic field, the surface energy density can be calculated by dividing the total energy density by the surface area of the wire, which can be found using the formula for the surface area of a cylinder (2πrL, where r is the radius and L is the length). Similarly, for the electric field, you can calculate the surface energy density by dividing the total energy density by the surface area of the wire, which can be found using the formula for the surface area of a circle (πr^2).

So, using the values you have already calculated, the surface energy density for the magnetic field would be 8.1 X 10^-6 J/m^2 and for the electric field, it would be 77.1 J/m^2. It is important to note that these values may vary slightly depending on the exact dimensions of the wire, so it is always a good idea to double-check your calculations with the correct formulas.
 

1. What is energy density?

Energy density is a measure of the amount of energy stored in a given space or system. It is typically expressed in units of joules per cubic meter (J/m3) for magnetic energy density and joules per cubic meter (J/m3) for electric energy density.

2. How is energy density related to magnetic and electric fields?

Energy density is directly related to the strength of the magnetic and electric fields in a system. The higher the strength of these fields, the higher the energy density will be. This means that increasing the strength of the fields will result in a higher energy density.

3. What factors affect the energy density of a magnetic or electric field?

The energy density of a magnetic or electric field is affected by several factors, including the strength of the field, the permeability or permittivity of the material it is passing through, and the size and geometry of the system. Additionally, the energy density can also be affected by external factors such as temperature and pressure.

4. Why is energy density important in scientific research?

Energy density is an important concept in scientific research because it allows scientists to understand and predict the behavior of magnetic and electric fields in various systems. It is also a crucial factor in the design and development of technologies that utilize these fields, such as motors and generators.

5. How is energy density calculated?

Energy density can be calculated by dividing the total energy of a system by its volume. For example, the energy density of a magnetic field can be calculated by dividing the magnetic energy by the volume of the space it occupies. Similarly, the energy density of an electric field can be calculated by dividing the electric energy by the volume of the space it occupies.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
724
  • Introductory Physics Homework Help
Replies
1
Views
742
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
665
  • Introductory Physics Homework Help
Replies
5
Views
811
  • Introductory Physics Homework Help
Replies
3
Views
136
  • Introductory Physics Homework Help
Replies
26
Views
474
  • Introductory Physics Homework Help
Replies
1
Views
716
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top