Magnetic flux rule for calculating motional EMF

[Pushoam]

Homework Statement

Griffith ( Introduction to electrodynamics , 3 ed.)
says in Problem 7.9:
An infinite number of different surfaces can be fit to a given boundary line, and
yet, in defining the magnetic flux through a loop, Φ = ∫B.da, I never specified the particular
surface to be used. Justify this apparent oversight.

Homework Equations

The Attempt at a Solution

I have always taken that surface which is perpendicular to the B and contains the loop as its boundary.
I guess that Griffith doesn't define the surface because what we want to calculate is
ε = - dΦ/dt and dΦ/dt doesn't change w.r.t. surface.

But ,I don' t know in which direction should I think?

In page no.296 , Griffith says
Apart from its delightful simplicity, it has the virtue
of applying to non- rectangular loops moving in arbitrary directions through non- uniform
magnetic fields; in fact, the loop need not even maintain a fixed shape.

But, won't calculating Φ for non- rectangular loops and non uniform magnetic fields itself be hard?

 

[cnh1995]

I guess that Griffith doesn't define the surface because what we want to calculate is
ε = - dΦ/dt and dΦ/dt doesn't change w.r.t. surface.

Yes. Because the small surface in which the flux is changing is contained in the given boundary line.

 

[Pushoam]

Because the small surface in which the flux is changing is contained in the given boundary line.

This I didn't get.
Will you please illustrate it?

 

[cnh1995]

This I didn't get.
Will you please illustrate it?

If I got the question correctly:

Consider a big metal ring S , w.r.t. which you want to calculate dΦ/dt.

Suppose there are 100 different small rings contained in the ring S and in one of them, there is a changing flux. Call this ring S'. Now the emf induced in S' is equal to the emf induced in S, since S' is contained in S.

But I'm not sure if I got the original problem statement correctly.

 

[Pushoam]

I think the question means something else.
Consider a ring shaped loop.
Consider two surfaces:
One could be a surface of the disk bounded by the ring.
Another could be a hemispherical surface like a bowl.
What the question says that while applying flux rule (or may be calculating flux), it doesn't matter whether I calculate flux through disk surface or hemispherical surface.
This I have to prove.

 

[cnh1995]

What the question says that while applying flux rule (or may be calculating flux), it doesn't matter whether I calculate flux through disk surface or hemispherical surface.

Well in that case, you need to consider the fact that all the field lines that pass through the circular base of the hemisphere also pass throgh the curved surface of the hemisphere. So it doesn't matter which surface you choose.

 

[Pushoam]

Well in that case, you need to consider the fact that all the field lines that pass through the circular base of the hemisphere also pass throgh the curved surface of the hemisphere. So it doesn't matter which surface you choose.

I want to show it mathematically.
How to do this ?

 

[cnh1995]

I want to show it mathematically.
How to do this ?

Sounds like you are talking about Stoke's theorem. Have a read.
https://www.google.co.in/url?sa=t&s...ggvMAM&usg=AFQjCNEIedKZwC-NJnbSNgpG2PwSlrQIdw

 

[Pushoam]

 

[cnh1995]

View attachment 204909 View attachment 204910

There is no mistake. The minus sign shows up because you are considering a closed surface, where incoming flux (through the circular base) is equal to the outgoing flux (through the curved surface). If incoming flux is positive, outgoing flux has a negative sign, making their total zero.

 

[Pushoam]

The minus sign shows up because you are considering a closed surface, where incoming flux (through the circular base) is equal to the outgoing flux (through the curved surface).

Thank you.
Now I got it.
While applying Green's theorem,
the positive direction of da for curved surface is $\hat r$ .
But in the second case , for calculating flux through the curved surface when the current is flowing through the ring, the positive direction of da for curved surface is -$\hat r$ (as if I approximate the curved surface by a plane surface then the positive direction of da for this surface is $\hat z$ which is possible only if the positive direction of da for curved surface is $-\hat r$) .

Hence, flux through the curved surface in the first case i.e. Green's theorem is negative of the flux through the curved surface in the Second case.
Hence, using the above fact and eq.(4), eq.(6) could be shown.

Generalizing it, magnetic flux through all different surfaces bounded by the same boundary line is same.

Is this correct?

 

[cnh1995]

Generalizing it,magnetic flux through all different surfaces bounded by the same boundary line is same.

Yes. All that goes in, comes out.

In Faraday's law, the flux is calculated w.r.t. open surface and hence, the integral of B.da is not zero.

 

[Pushoam]

The first question is solved, but the second still has to be solved.

In page no.296 , Griffith says
Apart from its delightful simplicity, it has the virtue
of applying to non- rectangular loops moving in arbitrary directions through non- uniform
magnetic fields; in fact, the loop need not even maintain a fixed shape.

But, won't calculating Φ for non- rectangular loops and non uniform magnetic fields itself be hard?

 

[cnh1995]

But, won't calculating Φ for non- rectangular loops and non uniform magnetic fields itself be hard?

It is hard.
I am not sure if I got your question correctly this time as well.
Could you elaborate?

 

[Pushoam]

Could you elaborate?

What I have understood from the Griffith's book is:
Griffith is saying that emf can be calculated (using flux rule ) even when
(1) the loop is non- rectangular
(2)the magnetic field is varying
(3)the loop is moving in any arbitrary direction and
(4) the shape of the loop doesn't remain fixed

But, I, too, feel that it is hard( can be almost impossible in some cases,too) to calculate flux in any of these four cases.

 

[cnh1995]

What I have understood from the Griffith's book is:
Griffith is saying that emf can be calculated (using flux rule ) even when
(1) the loop is non- rectangular
(2)the magnetic field is varying
(3)the loop is moving in any arbitrary direction and
(4) the shape of the loop doesn't remain fixed

Griffith is assuming that you have all the necessary information (otherwise what's the point of creating and solving such exercise?). For example, for a loop increasing in area in a magnetic field, emf can be calculated if you know the rate of increase in area and the variation of the magnetic field. It might be harder because of the higher math involved, but with sufficient information and correct math, none of the four cases is impossible.

 

[Pushoam]

none of the four cases is impossible.

o.k. thank you.

 

[cnh1995]

o.k. thank you.

You're welcome!

It would be better if you asked such conceptual questions in General physics or Classical physics forums. You could get way better and insightful answers there from our experts.

 

[rude man]

Green's theorem is just the special case of Stokes's theorem applied to a two-dimensional figure. At least that's the consensus. So I have a problem with the wording in the graphic of post 9.

 

[Pushoam]

Green's theorem is just the special case of Stokes's theorem applied to a two-dimensional figure

pictures from page no.29 and 34, chapter 1 , Introduction to electrodynamics,3 ed.

I have never read that Green's theorem is a special case of Stokes' theorem. Will you please give me a reference (to read)?

 

[cnh1995]

pictures from page no.29 and 34, chapter 1 , Introduction to electrodynamics,3 ed.
View attachment 204947
View attachment 204949

I have never read that Green's theorem is a special case of Stokes' theorem. Will you please give me a reference (to read)?

There are many articles available on the internet. This is one of them.
https://www.google.co.in/url?sa=t&s...ghHMAs&usg=AFQjCNH7m-8YDD_UvhxVmx_gxWqlcHVZhw

 

[Pushoam]

Well, then what I am applying is Gauss' theorem, not Green's theorem.'
And Gauss' theorem is not a special case of stokes' theorem. Right?

 

[rude man]

I have never read that Green's theorem is a special case of Stokes' theorem. Will you please give me a reference (to read)?

https://en.wikipedia.org/wiki/Green's_theorem
Read top paragraph.

There are really two theorems, quite different from each other. I call them the Divergence theorem and the Stokes theorem. "Green's Theorem" seems to lead to all sorts of confusion; in post 20 "Green's Theorem" even refers to the Divergence Theorem. A legitimate synonym for the Divergence Theorem is "Gauss's Theorem". In electrostatics it's often used to mean the Divergence Theorem as applied to one of Maxwell's equations: ∇ ⋅ D = ρ which in integral form becomes
∫∫D⋅dA = ∫∫∫(∇⋅D)dV = ∫∫∫ ρ dV = Q_free
and I hope you're sufficiently familiar with these symbols.

 

[rude man]

Well, then what I am applying is Gauss' theorem, not Green's theorem.'

No. You need Stokes's theorem as post 8 first mentioned.

And Gauss' theorem is not a special case of stokes' theorem. Right?

Right. See my previous post ←

 

[Pushoam]

No. You need Stokes's theorem as post 8 first mentioned.

But I have already solved the problem using Gauss' theorem i.e. Divergence theorem.
Isn't this o.k.?

 

[cnh1995]

Well, then what I am applying is Gauss' theorem, not Green's theorem.'
And Gauss' theorem is not a special case of stokes' theorem. Right?

I think what you used is one of the four Maxwell's equations. Gauss divergence theorem contains a volume integral, which I don't see in your equation.

 

[Pushoam]

In coming from eqn 1 to 2, I have used GAuss' theorem.
∫∇.B dΓ = $\oint_S$B.da = 0

 

[rude man]

But I have already solved the problem using Gauss' theorem i.e. Divergence theorem.
Isn't this o.k.?

I think you need to quote problem 7.9 for us verbatim to resolve this issue. I agree that what you wrote is correct but I wonder if it answers the question directly.

 

[Pushoam]

I think you need to quote problem 7.9 for us verbatim to resolve this issue. I agree that what you wrote is correct but I wonder if it answers the question directly.

Griffith ( Introduction to electrodynamics , 3 ed.)
says in Problem 7.9:
An infinite number of different surfaces can be fit to a given boundary line, and
yet, in defining the magnetic flux through a loop, Φ = ∫B.da, I never specified the particular
surface to be used. Justify this apparent oversight.

This is the statement of the problem given in the book. I am attaching the image.

 

[rude man]

I am attaching the image.
View attachment 205011

I do not see any image?
Oh, you mean the statement of the problem. No graphic?
The problem includes the statement " ... I never specified the particular surface to be used ... ". Please include the statement referred to herein. "Problem 7.9" is obviousy a continuation either of a previous problem or some text.

 

[Pushoam]

The above thing is the image.

Please click it.

 

[rude man]

The above thing is the image.

Please click it.

The problem includes the statement " ... I never specified the particular surface to be used ... ". Please include the statement referred to.

 

[Pushoam]

I attached the whole section of the book from which the question is asked.The definition of flux is given in page no. 295, first attachment.

 

[rude man]

I attached the whole section of the book from which the question is asked.The definition of flux is given in page no. 295, first attachment.

OK, thanks. I know the Stokes theorem is appropriate here, but I haven't decided yet on whether the way you invoked the Divergence theorem isn't also OK. That's my answer for the moment & I'll try to get back a bit later with any new views.

 

[Pushoam]

I know the Stokes theorem is appropriate here,

How can one apply stokes' theorem to answer this question ?
Will you please give me some hint?

 

[rude man]

How can one apply stokes' theorem to answer this question ?
Will you please give me some hint?

After looking at your pdf files I see that neither theorem is appropriate for proving that the shape of the surface enscribed by a contour (a loop) is immaterial. I have to apologize to you for going in that direction for so long but i did need to see those pdf pages.

In fact, either theorem requires the inclusion of a maxwell relation. And here's the problem with that: when dealing with moving media such as the loop of fig. 7.13 the maxwell relations are often irrelevant! The author himself points that out (p. 298 lines 8 and 9).

So, bottom line, I conclude that neither the Stokes nor the Divergence theorem is apposite to proving what he seems to be referring to. Referring again to fig. 7.13, the emf is generated differentially for every segment of the loop dl, so the attached surface is immaterial. The loop of fig. 7.13 is an example of where what I call the "Blv law" is the correct law to invoke, not any of the four maxwell relations: d(emf) = B⋅(dl x v) = (v x B)⋅dl. And so the total emf around the loop is just ∫(v x B)⋅dl. The shape of the surface has nothing to do with this integral!

As an example of where you luck out with maxwell is fig. 7.16. In this case emf = - dΦ/dt (based on maxwell's ∇ x E = - ∂B/∂t plus Stokes) happens to be correct but safer is to use the BLv law: emf = Blv based on the Lorentz law F = qv x B.