# Electromagnetic energy of a long wire

1. Dec 10, 2013

### hokhani

When we have a current in a long wire, what is the kind of total energy of the system? We do have an electric field inside the wire and a magnetic field outside it. I think the electric field has an energy, say E, and it is only this energy which appears also as magnetic energy and we don't have to take the sum of electric energy and magnetic energy as the total energy. Am I right?! Could anyone help me please?

2. Dec 10, 2013

### Astrum

Do you mean if there's a current through the wire? Or are you talking about a line charge and an external magnetic field?

3. Dec 10, 2013

### hokhani

yes.

4. Dec 10, 2013

### Astrum

You're really asking about the energy stored in E-fields and B-fields. The work to create a charge distribution (static) is $W_e = \frac{\epsilon _0}{2} \int E^2 d\tau$ and to create a magnetic field you need to go against the back emf, so the work is $W_m = \frac{1}{2\mu _0}\int B^2 d\tau$.

At this point we can see that $$U_{em} = \frac{1}{2} \int \left( \epsilon_0 E^2 + \frac{1}{\mu _0} B^2 \right) d\tau$$

Of course, the more complete and general expression is[/PLAIN] [Broken] Poynting's Theorem.

You can look up the details and the full deriviation, but the general idea is that we rewrite the Lorenz force as work form. $\mathbf F \cdot d \mathbf l = q(\mathbf E + \mathbf v \times \mathbf B ) \cdot \mathbf v = \mathbf E \cdot \mathbf v dt$ and rewrite it as $\frac{dW}{dt} = \int (\mathbf E \cdot \mathbf J ) d\tau$

From that point you need to play with vector identities and Faraday's law. It's a little bit messy.

Last edited by a moderator: May 6, 2017
5. Dec 12, 2013

### hokhani

Thank you. By this you imply that once the circuit becomes open and the current starts, some of electrical energy is transferring to magnetic energy so there is an opposition against the increasing current. But after a while that current becomes constant, we would have total energy as sum of electric and magnetic energy. Don't you?

6. Dec 12, 2013

### Astrum

The energy in $\mathbf B$ comes from having to do work against the back emf, this is work that needs to be done by the battery.

All the work done here is from the battery (or source of the current) rather than the fields themselves. Or at least that's the way I understand it.

7. Dec 12, 2013

### Ethan0718

1.We also have electric field outside the wire.

Do you know the potential difference? Roughly speaking, you can draw the electric lines from high potential to low potential in space.

2.As you've said, we have magnetic field outside the wire.

So, we have to take the sum of electric energy and magnetic energy as the total energy outside the wire if you just want to calculate the total energy. Perhaps the transmission power is also something you want to know. Then, you can calculate it outside the wire by poynting vector.

3.Enery is transmitted in space, rather than in circuit.

Please see this circuit simulation. I think this will help you!

http://www.oberlin.edu/physics/dstyer/CircuitSurveyor/

8. Dec 13, 2013

### FarticleFysics

When you say back EMF do you mean the field opposite of the induced electrical field from the magnetic field, or the induced field itself?

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