# Homework Help: Magnetic flux rule for calculating motional EMF

1. Jun 4, 2017

### Pushoam

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

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2. Jun 4, 2017

### cnh1995

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

3. Jun 4, 2017

### Pushoam

This I didn't get.

4. Jun 4, 2017

### cnh1995

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.

5. Jun 4, 2017

### 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.

6. Jun 4, 2017

### cnh1995

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.

7. Jun 4, 2017

### Pushoam

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

8. Jun 4, 2017

9. Jun 5, 2017

### Pushoam

10. Jun 5, 2017

### cnh1995

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.

11. Jun 5, 2017

### Pushoam

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?

12. Jun 5, 2017

### cnh1995

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.

13. Jun 5, 2017

### Pushoam

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

14. Jun 5, 2017

### cnh1995

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

15. Jun 5, 2017

### Pushoam

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.

16. Jun 5, 2017

### cnh1995

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.

17. Jun 5, 2017

### Pushoam

o.k. thank you.

18. Jun 5, 2017

### cnh1995

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.

19. Jun 5, 2017

### 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.

20. Jun 5, 2017

### Pushoam

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)?

21. Jun 6, 2017

### cnh1995

22. Jun 6, 2017

### 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?

23. Jun 6, 2017

### rude man

https://en.wikipedia.org/wiki/Green's_theorem

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 = Qfree
and I hope you're sufficiently familiar with these symbols.

24. Jun 6, 2017

### rude man

No. You need Stokes's theorem as post 8 first mentioned.
Right. See my previous post

25. Jun 6, 2017

### Pushoam

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