Physics question (university)?
Can someone please tell me if this approach is correct? It seems too simplistic to me.
If it is wrong, I would appreciate some advice.
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No, in my opinion that can't be right. Current^2*length doesn't have the dimension of energy, or am I missing something?
You have to take the magnetic field energy into account. When the currents flow in the same direction, the fields add up; when the currents flow in opposite directions, the field is reduced.
With the energy density of the magnetic field
is the work to be done
(B1, B2=fields inside the coils, V=volume of the coils). In the first case, energy is released (W<0), in the second case, work must be performed (W>0).
Thank you for the quick, specific answer. It helped me a lot.
Sorry, I didn't check what L meant and that your solution was almost correct…
I think L means inductance.😉
Ja
Ah yes, of course, I could have figured that out myself… The question then is which inductance is meant, since the coils are not the same. Or at most, the mutual inductance…
@Luca0533
Unfortunately, you divided by a factor of 2. That was the mistake.
If so, you have to take the mutual inductance M into account: the energy difference is contained in the mutual inductance M.
Assuming ideal coupling k=1 (no stray field),
The difference we are looking for is therefore (note: no factor 1/2)
This is of course the same as from , just a bit more compact.
Unfortunately, you incorrectly divided by 2. Therefore, your difference is too small by a factor of 2.
What confuses me a bit afterwards: in the above case, there is an attractive force when the currents in the coils flow in opposite directions (or am I seeing this wrong?).
However, with two thin coils or simple current loops, the force is attractive if the currents flow in the same direction, analogous to the case of parallel, straight conductors:
https://www.youtube.com/watch?v=FLWgm_j0XZc
Which, considering the Lorentz force, is immediately obvious. I'm wondering in retrospect whether it's sufficient to consider only the field energies in the above problem.
See also Griffits , page 373 (8.3): This article discusses the very old paradox in detail. More energy is required to increase the magnetic field energy than is released by mechanical work. The same effect also occurs in capacitors, but is more well-known.
Of course, there is also electrical energy. When the loops approach each other, mechanical energy is released, even though the field energy stored in the field increases. Therefore, there must be additional energy: this results from the voltage generated during the approach. The voltage is polarized in such a way that the current-supplying source supplies energy. Thus, everything is fine again.
The easiest way to see this is through the equations:
U1 = L1*I1' + M*I2' + I2*M'
If I1 and I2 remain the same and M increases due to the approach, then
U1 = I2*M' > 0, and thus energy is drawn from the current source. The energy balance is then correct again.
When the coils, which carry the same current, are pulled apart, more field energy is released than the mechanical work required. This energy flows into the power sources.
In my experience, there are very few people who rack their brains over something like this and actually understand the question. You're one of them 👍
The question is what is meant by "work done": the electrical energy supplied, the mechanical energy released, or the difference between the two. If mechanical work is meant, then it is clearly negative, since there is an attraction. Nevertheless, electrical energy is required from the sources, as the energy stored in the field increases.
The key to understanding is Eq. 8.43 in Griffiths (I wrote it above)
Yes, exactly, you got it.
Because of
U1 = I2*dM/dt
U1 dt = I2*dM
I1*U1*dt = I1*I2*dm
Each of the two sources must therefore supply the electrical energy
W(source) = I1*I2*M
As we know, the contribution
W(field) = I1*I2*M
into the coupled field.
The difference is due to mechanical work gained
W(mech) = -I1*I2*M compensated.
This is exactly what Griffiths described in his book.
OK, thanks, then it's comparable to a parallel-plate capacitor whose plates are connected to a voltage source. The plates attract each other, yet the field energy between the plates increases as the distance decreases. The voltage source provides twice the energy released as mechanical work.
But if that's the case, how can one calculate the work to be done using only the field energy? Or is this not just mechanical work?