Lenz's Law Question

1. Jul 19, 2011

krw

I have a question regarding Lenz's Law. In many of the passages that I read about Lenz's Law, I see this: "If the magnetic field of current i1 induces another current, i2, the direction of i2 is opposite that of i1."

I don't think this is always true, though. For example, let's assume we have two straight conductors in close proximity. One of these conductors has an AC current flowing through it. For about half of each cycle, the magnitude of the current is decreasing. During this time, the magnetic field produced by that conductor will be in a given direction, but the change in magnetic field should be in the opposite direction. (The magnetic field is "decelerating")

Because the magnetic field of an induced current must always oppose the change in the applied magnetic field, the magnetic field of the second conductor should be in the same direction as the magnetic field of the first conductor. This means that their currents should also be in the same direction.

So, why is there a discrepancy?

2. Jul 20, 2011

gerbi

Firstly, Len'z law isn't like : "If the magnetic field of current i1 induces another current, i2, the direction of i2 is opposite that of i1." It's about the magnetic fields, not currents.

Secondly, You treat this like some simple stuff.. i1 is opposite to i2.. what about inductance of i2 circut? Current induced in circut depends on voltage (described by Faraday's Law) but also on impedance (it's real and imaginary part) which changes the phase of current. For perfect inductance it will be -90 deg. Then the currents (i1 and i2) are perfect opposite.
You have to consider losses (if resistance in secondary wire is present). This will create some field changes.

It's not that easy..

3. Jul 20, 2011

krw

I never said that was Lenz's Law. You need to read more carefully.

I agree that the currents will be 90 degrees out of phase, but that does not equate to "perfectly opposite currents". Opposing currents would be 180 degrees out phase. Equal magnitudes, opposite signs.

I just thought of a better way to phrase this: If the strength of a current in a wire is decreasing, the strength of that wire's magnetic field is also decreasing. Any current induced by that magnetic field should flow in a direction such that IT'S magnetic field opposes the change in the original magnetic field.

The change in the original magnetic field is opposite its actual direction (because it is decreasing), but the opposite of THAT is the same direction as the original magnetic field.

So the new magnetic field is in the same direction as the original magnetic field, meaning the currents must be in the same direction. This seems to be supported by this passage from Wikipedia: "However, when a varying field is falling in strength, the induced field will be in the same direction as that originally applied, in order to oppose the decline."

The article I got that from is about eddy currents, but the same principle should apply.

My question still stands: why is there this discrepancy? Why am I reading that a current induced by another current (by means of its magnetic field) always opposes the original current, even though that seems to be false?

4. Jul 20, 2011

gerbi

ahh.. c'mon. Any mathematical language ?

You create current in a wire (assume that's excitation):
i(t)=sqrt(2)*1*sin(omega*t)
Field generated by this current will be, generally, almost the same by it's form (Biot-Savart). It will be distributed around wire with different amplitude but it will change the same as current in time with rate of (omega*t). Then is the flux (phi).. by integration B over S plane.

Faraday's Law is next. e=-(dphi/dt). You get voltage induced in the second wire. It's phase shifted by 90 degrees comparing to the flux->current excitation. Assume there is no resistance in second wire.. phase shift for another 90 degrees.. means 180 deg in compare to exc.. You won't get the same amplitude (it's not a perfect magnetic coupling).
Correct me if I'm wrong, I'm not a guru.

Try to simulate this with some field software (FEM maybe).. or mathcad at least..

There is no false.. that stands for a long time and believe me, someone would noticed it's wrong if it would be ;)

EDIT: and one more thing, You treat Len'z law like some big picture.. look at Faraday's law, Len'z is included there by the "-"

Last edited: Jul 20, 2011
5. Jul 20, 2011

Studiot

I have not seen exactly this question in textbooks so it is worth examining further.

With references to the sketches.

I have drawn three sketchs showing the situation around wire A and then placed wire B alongside as FIG4. Wire A with the driving current is shown by a single line on the vertical axis. Wire B, with the induced voltage (current) is shown as a double line.

Consider an instantaneous snapshot with the current travelling up the page as in FIG1.

This creates rings of magnetic flux encircling the wire as shown in 3D in FIG2. Note that the direction of the flux reaches a maximum, passes through zero, reverses and reaches a maximum in the other direction in phase with the driving current waveform.

This results in a graph of magnetic flux as shown dashed in FIG2 and repeated for convenience in FIG3. Note these rings are at right angles wire A and therfore perpendicular to a second wire set parallel to A. So they are correctly oriented to induce a voltage along the second wire.

But what of the induced voltage?

The combined Faraday's and Lenz laws tell us

$$V = - \frac{{d\phi }}{{dt}}$$

So examining FIG3

$$\frac{{d\phi }}{{dt}} = \max$$ at the zero points of the flux waveform.

Further the minus sign in the equation says that when the flux waveform is positive going (increasing) we can expect the induced voltage to be negative (in the opposite direction) to the inducing voltage (current).

It is very important to realise that this opposite direction nothing to do with phase.

There are only two directions, up along the wires and down along the wires.

These are opposite and it is this relationship that derives directly from the minus sign the confirms the induced voltage is opposed to (=in the opposite direction to) the driving waveform.

There is also a phase difference of 90 degrees between the waveforms caused by the fact the induced wavorm is the slope or derivative of the driving waveform.

Since the the start of the induced positive excursion takes place 90 degrees after the start of the driving positive excursion it is lagging by 90 degrees.

Both these relationships are shown in the juxtaposition of FIGs 3 & 4.

go well

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6. Jul 20, 2011

krw

Thank you for the detailed response. You communicated more eloquently what I was trying to say.

So the induced current does in fact lag the driving current by 90 degrees, which is evident if you compare FIG1 to FIG4 like you said.

In the time domain, this translates to currents that flow in opposite directions when the driving current is increasing in magnitude, and to currents that flow in the same direction when the driving current is decreasing in magnitude.

In other words, a driving current and induced current DO NOT always flow in opposite directions. Please correct me if I'm wrong.

7. Jul 21, 2011

gerbi

hmm... Studiot, does this theory fits to all aplications ? What kind of mathematical equation You used to get current (Fig 4) from voltage induced (it's the same shape as current on Fig 4 ?).

Last edited: Jul 21, 2011
8. Jul 22, 2011

krw

I thought of another important thing. The induced current in the second conductor would in turn induce a current in the first conductor. So there would be a mutual inductance effect that makes this problem much more complicated than I was making it out to be.

9. Jul 22, 2011

Evil Bunny

More importantly... How did you draw such perfect sine waves and circles? lol

10. Jul 25, 2011

gerbi

Studiot.. I'm really looking forward for your reply. If You don't know, if it was Your guessing, tell us so we won't be overthrowing the Lenz law!

Now that's the big picture. Problem is not so easy.. intrested in solving it ? Use some field simulation software.

Haha.. this question should be answered in the first place :)

Last edited: Jul 25, 2011
11. Jul 25, 2011

Bob S

You probably should look at this problem from the standpoint of transmission line theory. It is well known that the speed of light is c = 1/sqrt [μ0ε0], where μ0 an ε0 are the permeability and permittivity of free space. For this reason, the product of L and C in an air-gap transmission line is 1/c2, where L and C are the inductance and capacitance per unit length. So capacitance is as important as inductance in a circuit with two parallel conductors. You cannot have inductance without capacitance. Signals (which travel at the speed of light) in one conductor can induce signals in the other, and in certain cases (stripline or microstrip couplers), the induced signal is only in the direction opposite to the primary signal (directional coupler).

Bob S

12. Jul 25, 2011

Studiot

BobS has given the game away, although the clues were all there in my sketches.

I stated, very carefully, that the waves are the state along the wire in space at some instant in time.

So the vertical axes, as drawn, are distance not time. Phase is a statement about time not distance.

I also carefully described the wave in Fig1 as travelling up the page, and showed that the magnetic wave also travels up the page.

But Lenz law states that the induced voltage wave travels down the page ie in the negative direction or against the axis arrow. This is what the minus sign is all about in the differential equation I stated. Not only does it travel down the page but its maxima and zeros are shifted $\pi$/2 relative to the magnetic wave. This is what is meant by "in a direction to oppose the change". Note again that direction is a spatial characteristic and phase is a temporal one. Lenz law is silent about the phase, you have to work the phase out from the diagrams as I have done or the mathematics as below.

I cannot stress this too highly

In a direction to oppose the change does not mean 180 out of phase.

I stressed it and highlighted it, what more can I do?

Consider a general travelling (sinusoidal) wave

$$\varphi = a\sin \left[ {\frac{{2\pi }}{\lambda }\left( {y - vt} \right)} \right] = a\sin 2\pi \left( {\frac{y}{\lambda } - \frac{t}{T}} \right)$$

where phi is the magnetic flux, a is a constant, v is velocity (constant) , t is time, T is period, lambda is wavelength and I have used y for distance since I am using the y axis.

Then the induced voltage, V is

$$V = - \frac{{d\varphi }}{{dt}}$$

So differentiating the above expression

$$V = - \left\{ { - \frac{{2\pi a}}{T}\cos 2\pi \left( {\frac{y}{\lambda } - \frac{t}{T}} \right)} \right\} = - \left\{ { + \frac{{2\pi a}}{T}\sin \left[ {2\pi \left( {\frac{y}{\lambda } - \frac{t}{T}} \right) - \frac{\pi }{2}} \right]} \right\}$$

I'm sorry I tried to colour the sign and the phase change but my TEX is not up to that refinement so you will have to look carefully.

The negative sign is preserved through the calculation, outside the bracket and plays no part in determining the phase change, which appears as the final term in the line.

As to how the induced current in Fig4 is in phase with the induced voltage, well both wires must have resistance and Ohms law prescribes this.

Finally the elipses in Fig 2 were drawn with an ellipse template, using small sized ellipses. The 'sine' curves were approximated using the pointy ends of one of the larger ellipses in the template.

go well

13. Jul 26, 2011

gerbi

That's more detailed answer, appreciate it.

I have one more request. Applicate this theory to single phase two winding transformer. Calculate H field which will be applied to the core. What would be the phase relation between primary and secondary current ? Neglect no-load current.

14. Jul 26, 2011

Studiot

From the days before political correctness when Physics really was a practical subject.

This extract should answer your question and several others you have not yet asked.

go well

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15. Jul 26, 2011

gerbi

Now You know my point ? Look at the current vs time. What are their phases ?

16. Jul 26, 2011

Studiot

If the secondary is open circuit (= no load) there is no secondary current.

Why do you think the field developed around an infinite straight wire is applicable to a transformer?

17. Jul 26, 2011

gerbi

Ahh.. my poor english. You posted the answer. The currents are in opposite phase. Neglecting no-load current (no-load current flows in transformer primary winding when the secondary winding is open) supposed to make the calculations easier.

Now, that's universal theory ? I mean the no phase-shift between secondary current and voltage (neglecting inductance) ? Using Ohms law for this kind of circuits ?

18. Jul 26, 2011

Studiot

In the two straight wires example I am ignoring inductance - that is I am assuming the frequency is low enough to ignore inductance of a straight wire. That is why I simply use Ohm's law.

You should note that it is impossible to have two infinite wires in practice, there must be a wire carrying the drive current the other way to complete any real circuit.
You will note that textbooks always discuss loops as a result of this.

As regards the transformer you obviously have many loops and also (probably) a core concentrating the magnetic field. Note also that in a transformer great effort is made to keep the resistance of the windings as low as possible as resistance leads to wasted power as heat. The 'ideal' transformer has zero winding resistance - all the impedance offered is reactive (inductive).

The (quick) principle of operation of a transformer is as follows

An external alternating voltage applied to the primary circulates in the primary circuit as an alternating current in the primary.
This primary current produces an alternating flux in the core.
This flux links both the primary and secondary windings, the whole of the flux passes through every turn of both primary and secondary windings and generates the same EMF in accordance with lenz law in opposition to the applied voltage in each turn of both windings.

19. Jul 26, 2011

gerbi

Ignor it.. can we ? To make it simple - sure. To be strict with physics - no (Inductance means there is a field around that wire). But fine, that may be the way.

Guess that's obvious for everyone. The loop needs to be closed.

I wasn't talking about a ideal one. Just said to make the calcs simpler. You can neglect no-load current in some calculations - calcs like we talked here.

Still talking about an ideal transformer, right ? Then it's correct.

We can talk like this all night long Studiot. All I was hoping to hear was that You state some assumptions You used to write Your pictures. For someone who isn't familiar with all this it's tough to understand what he really sees. Someone may be mislead.. let's keep it clear.