Understanding Self-Inductance of a Coil: How Does it Happen?

[kira506]

Self inductance of a coil : how does it happen ? I've got some hardtime trying to imagine how it happens , the phrase which confused the most while studying was "each turn acts as a small magnet ",so why does it change polarities at the moment of switching the circuit on or off ? Also the whole concept of an induced current is kinda hard to understand , I can barely understand it ! I mean ,does the magneic field affect the free electrons of a conductor , so they move ?According to this idea , we can induce a current in all metals ... also, is there any other way to calculate self-inductance coefficient? One that doesn't include magnetic flux lines , magnetic in duction (B) ,cross sectional area of coil and time ,perhaps? Smethings which involves only current intensity and induced Emf ? If you answer these questions , I'd be forever indebted to you , afterall,my studies depend on this ,thank you so much in advance !

 

[Philip Wood]

Also the whole concept of an induced current is kinda hard to understand

But this is what you need to do! You need, imo, to get a feeling for electromagnetic induction before you tackle self inductance.

does the magneic field affect the free electrons of a conductor , so they move ?

Yes, that's true, but specifically: when the magnetic flux linking a circuit changes, an emf is induced in the circuit. This (the emf) is the work done per unit charge on charge which goes round the circuit, so can loosely be thought of as making the free electrons move round the circuit (if it is closed). You've now got to understand what it means to say that "the magnetic flux linking a circuit changes". Do you need help with this?

 

[kira506]

But this is what you need to do! You need, imo, to get a feeling for electromagnetic induction before you tackle self inductance.



Yes, that's true, but specifically: when the magnetic flux linking a circuit changes, an emf is induced in the circuit. This (the emf) is the work done per unit charge on charge which goes round the circuit, so can loosely be thought of as making the free electrons move round the circuit (if it is closed). You've now got to understand what it means to say that "the magnetic flux linking a circuit changes". Do you need help with this?

so I'm on the right track ? Thank you so much ! But I still can't fully understand how a coil can induce a current in itself , for example : inducing an emf in a direction opposite to the original ,and does the strenght of the induced emf have to be more than the original or does it depend on the factors we conclude from the equations ? Thanks a million in advance ! (and also for confirming my doubts about the electrons moving)

 

[kira506]

so I'm on the right track ? Thank you so much ! But I still can't fully understand how a coil can induce a current in itself , for example : inducing an emf in a direction opposite to the original ,and does the strenght of the induced emf have to be more than the original or does it depend on the factors we conclude from the equations ? Thanks a million in advance ! (and also for confirming my doubts

 

[Philip Wood]

First a point about current and emf. The change in magnetic flux linkage (due to the change in current in the coil) induces an emf in the coil. The current which flows is proportional to the resultant emf, that is the algebraic sum of the induced emf and any other emfs in the circuit.

If you're having difficulty with the idea of self-induction and self-inductance, the first thing to check is that you understand, and have done some problems about, the emfs induced in a circuit due to the flux linking the circuit changing. For example, can you do this question?

"A coil of 400 turns and diameter 0.050 m is placed in a magnetic field of flux density 0.12 T, so that the flux density is normal to the coil's cross-sectional area. What is the mean emf induced in the coil when it is turned through 90° (so that no flux is linked with it) in a time of 0.20 s."

If you can do this sort of problem, then we can safely go on to self inductance!

 

[alva]

Self inductance of a coil : how does it happen ?

Every circuit has "self inductance". I don't know why they call it "self inductance". It's just inductance.

A single straight wire has "self inductance" or inductance.

The fact is that a current produces a B ( magnetic ) field. An increasing B field produces an E ( electric ) field. And, as it happens, nobody knows why, the E is opposed to the current that produces the B field. And that is what we call "inductance".

A coil is no different from a single wire, it is just a way of making a high inductance with the same length of wire.

I wish all the mentors corrected me if there is something wrong.

 

[Dale]

I don't know why they call it "self inductance". It's just inductance.

To distinguish it from mutual inductance.

 

[alva]

To distinguish it from mutual inductance.

1) inductance 2) self inductance 3) mutual inductance

What is the difference between 1) and 2) ?

 

[Dale]

1) is unclear and 2) is clear.

 

[kira506]

First a point about current and emf. The change in magnetic flux linkage (due to the change in current in the coil) induces an emf in the coil. The current which flows is proportional to the resultant emf, that is the algebraic sum of the induced emf and any other emfs in the circuit.

If you're having difficulty with the idea of self-induction and self-inductance, the first thing to check is that you understand, and have done some problems about, the emfs induced in a circuit due to the flux linking the circuit changing. For example, can you do this question?

"A coil of 400 turns and diameter 0.050 m is placed in a magnetic field of flux density 0.12 T, so that the flux density is normal to the coil's cross-sectional area. What is the mean emf induced in the coil when it is turned through 90° (so that no flux is linked with it) in a time of 0.20 s."

If you can do this sort of problem, then we can safely go on to self inductance!

well dividing the diameter by 2 we gw the radius , then we get the original current in the coil from : MIN/2r (M is the permeability of the medium,air in this case,forgive me ,not the proper symbols on my mobile ) magnetic flux at the new position chages to 0 , at the fomer position(perpendicular ) it was max, delta magnetic flux would be (0-max) , since raye of change in (I) is directly proportional to rate of change in (magnetic flux linkage) delta I = 0 - Imax(original current) , we finally substitute in the equation : induced EMF =-L*delta(I)/delta(t) ,but L is not given XD so we substitute in (L*delta(I)=N*magnetic flux ) and we get the magnetic flux from B=AB cos theta ,L =400(0-2.35*10^-5/(0-5.96) =1.58*10^-3
ind.EMF= 0.047 volts , I'm sure the answers is wrong though XD , because I don't understand what's required , if the required is "at the instant of changing position" then its what I've written,but if its "detemining it at the new position* then it is zero because there's no linkage unless , changing its position by 90 degrees meant changing it to 180 position and in that case , I don't know what happens , so , do I pass ? :Df

 

[kira506]

Every circuit has "self inductance". I don't know why they call it "self inductance". It's just inductance.

A single straight wire has "self inductance" or inductance.

The fact is that a current produces a B ( magnetic ) field. An increasing B field produces an E ( electric ) field. And, as it happens, nobody knows why, the E is opposed to the current that produces the B field. And that is what we call "inductance".

A coil is no different from a single wire, it is just a way of making a high inductance with the same length of wire.

I wish all the mentors corrected me if there is something wrong.

XD I agree with Dalespam , to distinguish it from the mutual , the sick nomemclature , for example : why do they call uss "human beings" ? Is there a human not being or something ? Same idea
so the fact that the current is it still increasing produces an increasing magnetic flux which produce a rate of change in it momentarily producing induced e.m.f , thanks a lot , its getting clearer now .that reminds me of Kirchoff's law, that's my main problem . I want to know how a current can increase gradually although it supposed to pass as a wholein the circuit , the self inductance makes it increase or decrease gradually , thanks again

 

[Philip Wood]

kira123: Well done for trying the little question I set. It was about flux changes and induced emf, not about self-induction.

Here's how you do it.

The initial flux, $\Phi$, through the coil is
$$\Phi = BA$$
in which A =area of the coil = $\pi \ 0.025^2 \mbox{m}^2$, and B = initial flux density = 0.12 T.

Flux changes to zero in a time of 0.20 s.

So mean emf induced in one turn= $\frac{\Delta\Phi}{\Delta t} = \frac{BA - 0}{\Delta t} = \frac{\pi\ 0.025^2 \times\ 0.12}{0.20}$ = 1.18 mV.

But there are 400 turns in series with each other, so total emf induced = 400 x 1.18 mV = 0.47 V

No current will flow unless the ends of the coil are connected to something which will complete the circuit.

The idea of asking you to do this was to make sure that you were familiar with the idea of changing magnetic flux inducing an emf, and could do simple calculations - before embarking on the slightly tricky idea of self-induction.

If you've followed what I've done here, we can move on to self induction!

 

[Entanglement]

Self inductance of a coil : how does it happen ? I've got some hardtime trying to imagine how it happens , the phrase which confused the most while studying was "each turn acts as a small magnet ",so why does it change polarities at the moment of switching the circuit on or off ? Also the whole concept of an induced current is kinda hard to understand , I can barely understand it ! I mean ,does the magneic field affect the free electrons of a conductor , so they move ?According to this idea , we can induce a current in all metals ... also, is there any other way to calculate self-inductance coefficient? One that doesn't include magnetic flux lines , magnetic in duction (B) ,cross sectional area of coil and time ,perhaps? Smethings which involves only current intensity and induced Emf ? If you answer these questions , I'd be forever indebted to you , afterall,my studies depend on this ,thank you so much in advance !

It's the electromagnetic effect induced in the same coil to resist the change in current intensity producing it, connect a coil in series with a battery and a switch, when the current passes in the coil, a A strong magnetic field is formed because each turn acts as a short magnet whose magnetic flux intercepts the neighboring turns. When the coil circuit is switched off, the current decays gradually, and so does the magnetic field of the neighboring turns. Each turn cuts the decreasing lines of magnetic flux therefore An induced emf is generated due to self induction in the coil. According to lenz's rule the induced emf produces an induced current which flows in the same direction of the original current to keep the original current which delays the current decay

But when the circuit is switched on the same thing happens (change in magnetic flux which intercepts the coils)
But according to lenz's rule the induced emf produces a current which flows opposite in direction to original current to resist its increase, which delays the original current growth to the constant value which is calculated according to Ohms's law I=V/R

 

[kira506]

kira123: Well done for trying the little question I set. It was about flux changes and induced emf, not about self-induction.

Here's how you do it.

The initial flux, $\Phi$, through the coil is
$$\Phi = BA$$
in which A =area of the coil = $\pi \ 0.025^2 \mbox{m}^2$, and B = initial flux density = 0.12 T.

Flux changes to zero in a time of 0.20 s.

So mean emf induced in one turn= $\frac{\Delta\Phi}{\Delta t} = \frac{BA - 0}{\Delta t} = \frac{\pi\ 0.025^2 \times\ 0.12}{0.20}$ = 1.18 mV.

But there are 400 turns in series with each other, so total emf induced = 400 x 1.18 mV = 0.47 V

No current will flow unless the ends of the coil are connected to something which will complete the circuit.

The idea of asking you to do this was to make sure that you were familiar with the idea of changing magnetic flux inducing an emf, and could do simple calculations - before embarking on the slightly tricky idea of self-induction.

If you've followed what I've done here, we can move on to self induction!

Thanks , the best way to make someone understand something is providing them with an example with numbers , there's only one thing I don't get , why did you write the change in "BA" as "BA-0" ? Isn't it supposed to be the other way around , so that when we multiply with "-N" as in Faraday's law , the -ve sign would be canceled ?

 

[kira506]

It's the electromagnetic effect induced in the same coil to resist the change in current intensity producing it, connect a coil in series with a battery and a switch, when the current passes in the coil, a A strong magnetic field is formed because each turn acts as a short magnet whose magnetic flux intercepts the neighboring turns. When the coil circuit is switched off, the current decays gradually, and so does the magnetic field of the neighboring turns. Each turn cuts the decreasing lines of magnetic flux therefore An induced emf is generated due to self induction in the coil. According to lenz's rule the induced emf produces an induced current which flows in the same direction of the original current to keep the original current which delays the current decay

But when the circuit is switched on the same thing happens (change in magnetic flux which intercepts the coils)
But according to lenz's rule the induced emf produces a current which flows opposite in direction to original current to resist its increase, which delays the original current growth to the constant value which is calculated according to Ohms's law I=V/R

Yes ! I'm close to understanding , just could you please explain to me the part where "each turn acts as a magnet" ?I want to visualize it but can't seem to

 

[Entanglement]

Yes ! I'm close to understanding , just could you please explain to me the part where "each turn acts as a magnet" ?I want to visualize it but can't seem to

You have to understand the nature of electromagnetism of the circular loop first, we can consider a circular loop as two STRAIGHT wires connected to each other. Each wire carries the same current but in an opposite direction, to visualize the magnetic field of the circular coil we will deal with loop as two connected Straight wires carrying an opposite current. By applying ampere's right hand rule on each wire, it is obvious that both magnetic fields "arising from what we consider two straight wires " have the same direction at the center of the loop forming a nearly uniform magnetic field, and by applying the field direction on both sides of the loop, you will realize that each side has a different polarity "where the field goes out at one side and goes inside at the other side". That's why you have to put in your consideration that the magnetic loop is a DIPOLE magnet it acts exactly like the bar magnet as it has a south and a North Pole
Let's get back to the the inductance coil.
Solenoids consist of circular loops connected to each other " in which there is a separating distance between each loop " when the magnetic flux falls on the solenoid, each loop responds independently to the magnetic flux change where an emf is induced in each loop, you can visualize this as batteries connected in series in a circuit where you add up their emfs.When a current flows in the solenoid each loop becomes magnetized affecting the neighboring turns with its flux

 

[Philip Wood]

You are right. Change is always to be interpreted as (final value - initial value). This isn't an excuse, but I wasn't concerned with the direction of the emf.

Now for some self-induction… In the example, the change in flux was due to the coil changing its position in an external magnetic field. The same equation would have applied, and we'd have got the same answer, if the coil had stayed put, but we had reduced the external magnetic field from 0.12 T to 0 T in the same time (0.20 s).

Suppose, now that there is no external field, but the coil has a current flowing through it. The coil will produce its own field, and flux will be linked with it: the lines of flux are closed loops and will pass through the cross-section of the coil and will curl round outside the coil to complete the loops.

[Example a solenoid of length 1.00 m, with 2000 turns, carrying a current of 3.0 A will have a flux density inside it of $\mu_0 \times 2000 \times 3.0 =7.54$mT, so the flux passing through it will be BA = 1.51 x 10^-5Wb, if the cross-sectional area, A, is 2.0 x 10^-3 m². If the current falls to zero in 0.2 s, the induced voltage per turn will be 1.51/0.20 = 7.54 x 10^-5 V, so the total induced voltage will be 2000 x 7.54 x 10^-5 = 0.15 V]

In algebra - see of you can do this - the emf is given by

$$emf =\ –\frac{\mu_0\ n^{2} A}{L} \frac{\Delta I}{\Delta t}$$

$\frac{\mu_0\ n^{2} A}{L}$ is a constant for the solenoid called its self-inductance, L.

Thus $emf =\ - L\ \frac{\Delta I}{\Delta t}$.

This defines L. Short coils and flat coils have values of L which can't be calculated using $\frac{\mu_0\ n^{2} A}{L}$.

 

[Entanglement]

you have to understand the nature of electromagnetism of the circular loop first, we can consider a circular loop as two straight wires connected to each other. Each wire carries the same current but in an opposite direction, to visualize the magnetic field of the circular coil we will deal with loop as two connected straight wires carrying an opposite current. By applying ampere's right hand rule on each wire, it is obvious that both magnetic fields "arising from what we consider two straight wires " have the same direction at the center of the loop forming a nearly uniform magnetic field, and by applying the field direction on both sides of the loop, you will realize that each side has a different polarity "where the field goes out at one side and goes inside at the other side". That's why you have to put in your consideration that the magnetic loop is a dipole magnet it acts exactly like the bar magnet as it has a south and a north pole
let's get back to the the inductance coil.

 

[kira506]

You have to understand the nature of electromagnetism of the circular loop first, we can consider a circular loop as two STRAIGHT wires connected to each other. Each wire carries the same current but in an opposite direction, to visualize the magnetic field of the circular coil we will deal with loop as two connected Straight wires carrying an opposite current. By applying ampere's right hand rule on each wire, it is obvious that both magnetic fields "arising from what we consider two straight wires " have the same direction at the center of the loop forming a nearly uniform magnetic field, and by applying the field direction on both sides of the loop, you will realize that each side has a different polarity "where the field goes out at one side and goes inside at the other side". That's why you have to put in your consideration that the magnetic loop is a DIPOLE magnet it acts exactly like the bar magnet as it has a south and a North Pole
Let's get back to the the inductance coil.
Solenoids consist of circular loops connected to each other " in which there is a separating distance between each loop " when the magnetic flux falls on the solenoid, each loop responds independently to the magnetic flux change where an emf is induced in each loop, you can visualize this as batteries connected in series in a circuit where you add up their emfs.When a current flows in the solenoid each loop becomes magnetized affecting the neighboring turns with its flux

Thank you so much , that was beyond helpful , that really helped me understand why they act as small turns , but there's only one question concerning the circular loop which confuses me , about the field entering the loop at a point annd exiting it at another , can you please explain it to me ? (the fact that a loop acts as a shortbar magnet)

 

[kira506]

You are right. Change is always to be interpreted as (final value - initial value). This isn't an excuse, but I wasn't concerned with the direction of the emf.

Now for some self-induction… In the example, the change in flux was due to the coil changing its position in an external magnetic field. The same equation would have applied, and we'd have got the same answer, if the coil had stayed put, but we had reduced the external magnetic field from 0.12 T to 0 T in the same time (0.20 s).

Suppose, now that there is no external field, but the coil has a current flowing through it. The coil will produce its own field, and flux will be linked with it: the lines of flux are closed loops and will pass through the cross-section of the coil and will curl round outside the coil to complete the loops.

[Example a solenoid of length 1.00 m, with 2000 turns, carrying a current of 3.0 A will have a flux density inside it of $\mu_0 \times 2000 \times 3.0 =7.54$mT, so the flux passing through it will be BA = 1.51 x 10^-5Wb, if the cross-sectional area, A, is 2.0 x 10^-3 m². If the current falls to zero in 0.2 s, the induced voltage per turn will be 1.51/0.20 = 7.54 x 10^-5 V, so the total induced voltage will be 2000 x 7.54 x 10^-5 = 0.15 V]

In algebra - see of you can do this - the emf is given by

$$emf =\ –\frac{\mu_0\ n^{2} A}{L} \frac{\Delta I}{\Delta t}$$

$\frac{\mu_0\ n^{2} A}{L}$ is a constant for the solenoid called its self-inductance, L.

Thus $emf =\ - L\ \frac{\Delta I}{\Delta t}$.

This defines L. Short coils and flat coils have values of L which can't be calculated using $\frac{\mu_0\ n^{2} A}{L}$.

yeah , I've been bothered a lot by the fact that everytime I use the equation for selfinductance of coils ,it doesn't work with a solenoid! (thanks for the equation) , so sorry for causing you much trouble but one more question , what's the difference between short coils or flat ones and a solenoid ? Why won't the ordinary equation work out with the solenoids ( so sorry again)

 

[Philip Wood]

No trouble!

Solenoids are long coils. If the length is much greater than the diameter, then the flux density is pretty uniform across the cross-section and throughout the length (except near the ends). That's why there's such an easy formula for the self-inductance [Did you do the derivation suggested in my last post?]

'Flat' coils, and short coils don't have uniform fields inside their cross-sections: the flux density, B, varies from point to point. That's why it's difficult to work out their inductances, though formulae have been derived (as you can find by searching the web).

 

[kira506]

No trouble!

Solenoids are long coils. If the length is much greater than the diameter, then the flux density is pretty uniform across the cross-section and throughout the length (except near the ends). That's why there's such an easy formula for the self-inductance [Did you do the derivation suggested in my last post?]

'Flat' coils, and short coils don't have uniform fields inside their cross-sections: the flux density, B, varies from point to point. That's why it's difficult to work out their inductances, though formulae have been derived (as you can find by searching the web).

At that time while solving the problem I didn't use the derivation , it didn't come up to my mind and when I saw it , I didn't know how it came to be like that so I took it as it is , but thanks , now I understand , because B of solenoid is nearly uniform so we can use a simplified formula specifically to calculate its inductance , while we use the general forumla for the others because their B is not constant along their centers ,thank you so much , I'm really sorry for having troubled you much by asking numerous questions . Thanks a million again

 

[kira506]

View attachment 65312

Oh , now I get it ! You meant that the two poles are located along the flux in its center , I thought they were on the same plane as (the place where the concentric circles are crowded) thank you so much for the illustration , it was really helpful , thanks a million again !

 

[Entanglement]

it was really helpful , thanks a million again !

You're welcome !