Comparing self inductance of three solenoids

In summary, the homework statement is saying that the self inductance of a solenoid does not depend on the sense of the windings.
  • #1
Jahnavi
848
102

Homework Statement


three solenoids.jpg


Homework Equations

The Attempt at a Solution



Self inductance depends on the geometry of the solenoid . It doesn't depend on the sense of windings . So , L1 = L2 .

But I don't understand how to deal with superconducting wires .

Does a superconducting wire affect the self inductance of a solenoid ?
 

Attachments

  • three solenoids.jpg
    three solenoids.jpg
    51 KB · Views: 766
Physics news on Phys.org
  • #2
"Self inductance ... doesn't depend on the sense of windings"
What they say is, " the direction (presumably sense) of winding was reversed in each layer."

I agree with you that 100 turns CW will give the same inductance as 100 turns ACW.
But what about 50 t CW and 50 t ACW instead of 100 t in the same sense?
 
  • #3
Merlin3189 said:
But what about 50 t CW and 50 t ACW instead of 100 t in the same sense?

Sorry , I don't understand your question .
 
  • #4
Say S1 is 100 turns in a right handed sense and S3 is also 100 turns in a right handed sense.
But S2 is not 100 turns in a left handed sense. It is, maybe, 50 turns in a left handed sense, then 50 turns in a right handed sense - reversing in each layer. Or it could reverse after each 25, or after each 10. (It might complicate things if it reversed after each 30, or 20, or some number where the number of turns in each sense were different.)

So the question was, what inductance would you get if you wound 5o turns in a right handed sense and the next layer as 50 turns in a left handed sense?

To forestall another possible question: "In a right handed sense" could mean, looking at one end of the solenoid and choosing one end of the wire as the start, then the wire proceeds clockwise around the solenoid. LHS would then mean, looking at the same end and using the same starting point of the wire, you see it going anticlockwise around the solenoid.
As far as resistance goes, how do we model a real inductor?
Which parts of the model would change or stay the same in the three cases?
 
  • #5
Merlin3189 said:
As far as resistance goes, how do we model a real inductor?

By considering its resistance in series with its inductance .

Merlin3189 said:
Which parts of the model would change or stay the same in the three cases?

L3 is an ideal inductor with no resistance .

This is what I would like to understand . How does absence of resistance affect the inductance of the coil .

Answer given is option b) . How is L3 = 0 ?
 
  • #6
By considering its resistance in series with its inductance .
This is, sort of, what I expected. But I would say, I model a real inductor as an ideal inductance in series with an ideal resistance. (Thankfully leaving aside the capacitance for now. Real inductors are nasty little things!)
L3 is an ideal inductor with no resistance .
So L3 is an ideal inductance in series with a zero resistance.

So what are L1 and L2?

Then, what changes between the three? Obviously the resistance changes - we are told that! Why do you think the ideal inductance changes?
 
  • Like
Likes Jahnavi
  • #7
Just noticed your other comment, that option b) is correct. I don't understand how L3 can be 0, so you need some better help than mine.
 
  • Like
Likes Jahnavi
  • #8
Merlin3189 said:
I don't understand how L3 can be 0

Exactly :smile:

@kuruman , @Charles Link Could you please give your opinion .

@Merlin3189 and I agree that option a) is correct .But answer given is option b) .
 
  • #9
Well, I don't agree with a) either !

The point I was trying to make, is that the resistance does not affect the inductance. That you are right in attributing it to geometry here.

My take was that,
L1 = Lx H in series with R Ω
L2 = Ly H in series with R Ω
L3 = Lz H in series with 0 Ω
And that, agreeing with your original thought, Lx = Lz, because the geometry is the same, but L2 is different because the geometry is different.
 
  • Like
Likes Jahnavi
  • #10
I would go with (c), but they weren't very clear on how the direction is "reversed in each layer". Do they have an even number of layers, etc? with the magnetic fields from each layer canceling each other? ## \\ ## Meanwhile, inductance is defined as ## L=\frac{\phi}{I} ##. Clearly, it doesn't matter if the wire is resistive or is superconducting.
 
  • Like
Likes Jahnavi
  • #11
I agree with c) . Even if the opposing layers did not exactly cancel, that is the only answer with L1=L3≠L2
 
  • Like
Likes Charles Link
  • #12
It makes you wonder if the person who made up the problem has a misconception about the magnetic field from superconducting wire. Inside a superconductor, the magnetic field vanishes, (I believe that is called the Meissner effect), but it doesn't mean that you can't create a magnetic field by using superconducting wire in a solenoid. ## \\ ## Edit: See https://en.wikipedia.org/wiki/Meissner_effect
 
  • Like
Likes Jahnavi
  • #13
Merlin3189 said:
but L2 is different because the geometry is different.

L = μ0n2lA . How does sense of winding change the geometry of the coil ?

Merlin3189 said:
I agree with c) . Even if the opposing layers did not exactly cancel, that is the only answer with L1=L3≠L2

But how can L2 be 0 as given in the option ? It is definitely non zero .
 
  • #14
Charles Link said:
## \\ ## Meanwhile, inductance is defined as ## L=\frac{\phi}{I}##

Does flux have direction ? If not then how is L1 ≠ L2 ?
 
  • #15
Jahnavi said:
Does flux have direction ? If not then how is L1 ≠ L2 ?
The question isn't clear on what they mean by "reversed". (a) is certainly a possibility. In this case, the question is too imprecise to be worth spending a lot of effort to try to determine what is the correct answer to the question. I think we are all in agreement on our magnetism concepts in regard to the inductance from a solenoid. The question needs to be worded much better so that it doesn't lead to extra confusion for those who try to determine the answer.
 
  • #16
Flux certainly does have direction and sense.

You can imagine L2 as two solenoids with half the turns. One is energised to be NS and the other to be SN. the two fluxes cancel out.
When wound normally, the two fluxes are in phase, both NS or both SN.
 
  • Like
Likes Jahnavi
  • #17
Merlin3189 said:
Flux certainly does have direction and sense.

You can imagine L2 as two solenoids with half the turns. One is energised to be NS and the other to be SN. the two fluxes cancel out.
When wound normally, the two fluxes are in phase, both NS or both SN.
To reverse the direction, you would need to have a small section at the end that basically does a "U" before making the next layer. (You would need to fold the wire back on itself).
 
  • Like
Likes Jahnavi
  • #18
Merlin3189 said:
I agree with you that 100 turns CW will give the same inductance as 100 turns ACW.

Can't we just simply assume that L1 has clockwise windings and L2 has anticlockwise windings ?

Even though direction of magnetic field is opposite , magnitude of the ratio Φ/I will be same for L1 and L2 making them equal .
 
  • #19
solenoid2.png
 

Attachments

  • solenoid2.png
    solenoid2.png
    11.6 KB · Views: 388
  • Like
Likes Charles Link
  • #20
Jahnavi said:
Can't we just simply assume that L1 has clockwise windings and L2 has anticlockwise windings ?

Even though direction of magnetic field is opposite , magnitude of the ratio Φ/I will be same for L1 and L2 .
That's correct. The question isn't clear on what they mean by reversal. (Are all of the layers reversed from the first solenoid, or do the layers alternate in the second solenoid?) And clearly, the correct answer is not (b). I think they basically goofed in a couple of places.
 
  • Like
Likes Jahnavi
  • #21
Jahnavi said:
Can't we just simply assume that L1 has clockwise windings and L2 has anticlockwise windings ?

Even though direction of magnetic field is opposite , magnitude of the ratio Φ/I will be same for L1 and L2 .
If what you say were the right assumption, yes.
But it says the winding reverses on each layer, so they can't all be in the same direction.

The diagram I found isn't quite right. I'll look for another.
 
  • Like
Likes Jahnavi
  • #22
Charles Link said:
The question isn't clear on what they mean by reversal.

Merlin3189 said:
But it says the winding reverses on each layer

Do you agree that if the words "in each layer " are removed from the statement it would imply that the sense of windings is opposite in L1 and L2 ( if one is CW other is ACW ) ?

Would you then agree that option a) would be correct ?
 
  • Like
Likes Charles Link
  • #23
Yes.

I can't find a solenoid wound that way - I expect there is no use for one like that. The bifilar winding (intended for non-inductive wire-wound resistors) is more likely to give exactly equal and opposite inductances. If you wind one layer on top of the other, the top one has a slightly bigger area.
 
  • Like
Likes Jahnavi
  • #24
Glad that we all agree :smile:
 
Last edited:
  • Like
Likes Charles Link
  • #25
Jahnavi said:
Glad that we all agree :smile:

No ! Glad that both of you agree with me :wink:
Since you brought me into this discussion in post #8, can you clarify what y'all are agreeing with? I would say that ##L_2=0##. There is only one choice where that is the case.
 
  • #26
kuruman said:
Since you brought me into this discussion in post #8, can you clarify what y'all are agreeing with?

Please refer to post#22 .

kuruman said:
I would say that ##L_2=0##. There is only one choice where that is the case.

How ?

Is it because windings are reversed in each layer such that if first winding is CW , next is ACW , then CW , such that flux from any two neighbouring turns cancel each other . The net flux for the complete solenoid being zero , making L2 = 0 ?
 
  • #27
Jahnavi said:
The net flux for the complete solenoid being zero , making L2 = 0 ?
Imagine there are 100 turns, out of which 50 are CW and 50 are CCW. This means two 50-turn coils are connected in series-opposition.
While their individual "self-inductance" is L (not zero, as it does not depend on the sense of the winding), the effective inductance is zero (assuming perfect magnetic coupling).
The mutual inductance of the coils gets subtracted from the self inductance and the effective inductance is
Leff=L1self+L2self-2M.
M=(k)*√(L1selfL2self), where k is the coupling co-efficient.
If there is perfect magnetic coupling, k=1 and M=√(L1selfL2self)=L
Since M=L, Leff becomes 0.
I would go with a).

There is a phenomenon called 'the Meissner effect 'in superconductivity, but I don't think that is supposed to be used here. (Not really sure about this though.)
 
Last edited:
  • Like
Likes kuruman
  • #28
I see my blind spot now, (a) is the answer.
 
  • #29
Hi cnh1995 ,

cnh1995 said:
Since M=L, Leff becomes 0.
I would go with a).

Interesting !

You showed L = 0 and then chose option a) Did you mean option c) :smile: ?

What is the mistake in the reasoning in latter part of post#26 ?
 
  • #30
Jahnavi said:
You showed L = 0 and then chose option a) Did you mean option c) :smile: ?
No, a) it is:wink:.
cnh1995 said:
While their individual "self-inductance" is L (not zero, as it does not depend on the sense of the winding), the effective inductance is zero (assuming perfect magnetic coupling).

Jahnavi said:
What is the mistake in the reasoning in latter part of post#26 ?
The reasoning is fine, but it doesn't make L2 zero (L1,L2, L3are "self" inductances as per the problem statement).
 
  • #31
@cnh1995 When split into composite parts, ## L_2 ## can be composed of self inductances and mutual inductance, but in the final tally, the inductance of ## L_2 ## includes the mutual inductance of the component parts. ## L_2=0 ## with your computation.
 
Last edited:
  • #32
Charles Link said:
but in the final tally, the inductance of L2L2 L_2 includes the mutual inductance of the component parts. L2=0L2=0 L_2=0 with your computation.
Yes, that's why in #27, I said the "effective inductance" is zero (self minus mutual). But in the problem statement, it is mentioned that L2 is the "self" inductance of the coil, which is not zero.
 
  • Like
Likes Jahnavi
  • #33
cnh1995 said:
Yes, that's why in #27, I said the "effective inductance" is zero (self minus mutual). But in the problem statement, it is mentioned that L2 is the "self" inductance of the coil, which is not zero.
In that case, in the statement of the problem, the first time they mentioned ## L_2 ## they called it the "inductance of the coil", and the second time they called it the "self-inductance of the coil". This one is really splitting hairs with the use of the terms, when they themselves were rather careless in the use of the terms. I think they need to be much more clear in the statement of the problem.
 
  • Like
Likes Jahnavi, kuruman and cnh1995
  • #34
Thanks all for your precious inputs .
 
  • #35
Charles Link said:
In that case, in the statement of the problem, the first time they mentioned ## L_2 ## they called it the "inductance of the coil", and the second time they called it the "self-inductance of the coil". This one is really splitting hairs with the use of the terms, when they themselves were rather careless in the use of the terms. I think they need to be much more clear in the statement of the problem.
I agree. That's where I tripped at first.
 

Related to Comparing self inductance of three solenoids

1. What is self inductance?

Self inductance is a property of a circuit element, such as a solenoid, that describes the ability of that element to generate an electromotive force (EMF) in response to a changing current. It is measured in units of henries (H).

2. How is self inductance calculated?

Self inductance can be calculated using the formula L = μN²A/l, where L is the self inductance in henries, μ is the permeability of the material, N is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

3. What factors affect the self inductance of a solenoid?

The self inductance of a solenoid is affected by factors such as the number of turns, the cross-sectional area, the length, and the material of the solenoid. Additionally, the presence of a core material within the solenoid can also affect its self inductance.

4. How can the self inductance of three solenoids be compared?

The self inductance of three solenoids can be compared by measuring each solenoid's inductance using a multimeter or by calculating it using the formula mentioned in question 2. By comparing the values, one can determine which solenoid has a higher or lower self inductance.

5. Why is it important to compare the self inductance of three solenoids?

Comparing the self inductance of three solenoids can provide valuable information about the properties of each solenoid and how they may behave in a circuit. This can help in selecting the most suitable solenoid for a particular application and in understanding the behavior of the circuit as a whole.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
344
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
802
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
2K
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
2
Views
890
Back
Top