Does magnetic energy storage with iron cores result in greater energy recovery?

[Rework]

Hi - On this forum, dgreenheck on May 20, 2013 touched on this area. I will post my question in this manner:

Say one has a solenoid with a given number of turns; if the core is air, we know that if we run 10 amps through the circuit, the energy to charge a magnetic field to "B" Tesla is recovered when the magnetic field collapses. Now if that core is iron instead of air, we know the magnetic field is greater than the original "B" value (assuming no saturation), due to the permeability of iron. My question is: when that field collapses, does one recover the same amount of energy as before (with the air core), or more energy that is contained in the larger "B" field? It would seem against all rules of conservation of energy that one would get more energy back, but where does that extra field energy go? This is a puzzle to me? You can't get something for m\nothing!

 

[berkeman]

Could you post a link to the original thread?

As discussed at the start of your previous thread, it takes more energy to get the 2nd coil up to 10A.

 

[Rework]

Could you post a link to the original thread?

As discussed at the start of your previous thread, it takes more energy to get the 2nd coil up to 10A.

https://www.physicsforums.com/threads/magnetic-energy-of-solenoid-with-without-core.692789/

 

[Rework]

https://www.physicsforums.com/threads/magnetic-energy-of-solenoid-with-without-core.692789/

I thought it takes less H to get the same B with the steel core? B=MU X H

 

[berkeman]

I thought it takes less H to get the same B with the steel core? B=MU X H

But in your OP in this thread, I got the impression you were wanting to have 10A of coil current in both cases...

 

[Rework]

But in your OP in this thread, I got the impression you were wanting to have 10A of coil current in both cases...

Correct. So what I am not understanding is that apparently, even though the current is the same, it takes more energy in the case of steel, to get to full current. The B field is greater, and contains more energy when collapsing, so you get the energy back. So even though the amp-turns "H" are the same, the field B is greater in the case of steel. So H seems to be flakey.

 

[russ_watters]

Correct. So what I am not understanding is that apparently, even though the current is the same, it takes more energy in the case of steel, to get to full current. The B field is greater...

Current isn't energy. It takes voltage (EMF) to drive a current to "charge" the induction. The [integrated] voltage times current is the energy.
http://farside.ph.utexas.edu/teaching/302l/lectures/node103.html
(first few sentences)

So even though the amp-turns "H" are the same, the field B is greater in the case of steel. So H seems to be flakey.

It's not. You're equating "amp-turns" with energy and they aren't the same thing.

 

[Rework]

Current isn't energy. It takes voltage (EMF) to drive a current to "charge" the induction. The [integrated] voltage times current is the energy.
http://farside.ph.utexas.edu/teaching/302l/lectures/node103.html
(first few sentences)

It's not. You're equating "amp-turns" with energy and they aren't the same thing.

This is a statement from "milesyoung" in the 2013 thread: Think of it this way, it takes less effort (work) to increase the magnitude of B in a steel core solenoid than in an air core solenoid, so for the same magnitude of B, less potential energy is stored.

Current isn't energy. It takes voltage (EMF) to drive a current to "charge" the induction. The [integrated] voltage times current is the energy.
http://farside.ph.utexas.edu/teaching/302l/lectures/node103.html
(first few sentences)

It's not. You're equating "amp-turns" with energy and they aren't the same thing.

This is a statement from 'milesyoung' in the 2013 thread: "Think of it this way, it takes less effort (work) to increase the magnitude of B in a steel core solenoid than in an air core solenoid, so for the same magnitude of B, less potential energy is stored."
That's what I'm getting at -- it seems you get more B with less energy input.
In any case, my reason for the question was for the stored energy in a SMES. You've convinced me that an air core is the only way to go for a SMES (as far as we know today), because of the high Tesla. We will know more.

 

[russ_watters]

This is a statement from 'milesyoung' in the 2013 thread: "Think of it this way, it takes less effort (work) to increase the magnitude of B in a steel core solenoid than in an air core solenoid, so for the same magnitude of B, less potential energy is stored."
That's what I'm getting at -- it seems you get more B with less energy input.

Yes. Just like if you lift a lighter weight, it takes less force, less input energy - and therefore stores less potential energy - to lift it to the same height as a heavier weight.

In any case, my reason for the question was for the stored energy in a SMES. You've convinced me that an air core is the only way to go for a SMES (as far as we know today), because of the high Tesla. We will know more.

Someone besides me, with a better handle on the technology, will need to comment on the pros and cons and typical practices there.

 

[Rework]

Yes. Just like if you lift a lighter weight, it takes less force, less input energy - and therefore stores less potential energy - to lift it to the same height as a heavier weight.

Someone besides me, with a better handle on the technology, will need to comment on the pros and cons and typical practices there.

So can I conclude that you get back what you put in, whether it is air or steel? Apparently no one can explain in simple terms what is happening when you tickle iron with a small H field and get a large B magnetic field. Please excuse my BSME.