# Fardays Law and Stokes Law

1. Nov 13, 2008

### TFM

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

Faraday’s Law can be written as:

$$\oint_P \vec{E} \cdot \vec{dl} = -\frac{d}{dt}\Phi$$

Where $$\Phi$$ is the magnetic flux. Use Stokes’ theorem to obtain the equvilant Maxwell equation (i.e. Faraday’s Law in differential form).

2. Relevant equations

Stokes' Law:

$$\int_{\partial s}F \cdot ds = \int_P (\nabla \times F) \cdot da$$

3. The attempt at a solution

So far, I have:

$$\int_{\partial s}F \cdot ds = \int_S (\nabla \times F) \cdot da$$

$$\int_{P}E \cdot dl = \int_S (\nabla \times E) \cdot da = \frac{d}{dt}\Phi$$

Does this look like i'm doing it right?

TFM

2. Nov 13, 2008

### gabbagabbahey

Looks fine, except your missing a negative sign in yur last expression and the path integrals in Stokes Law are always closed path integrals : $$\oint_{\mathcal{P}}$$ ....Now use the definition of 'magnetic flux' through a surface.

3. Nov 13, 2008

### TFM

Re: Faraday's Law and Stokes' Law

So:

$$\oint_{P}E \cdot dl = \oint_S (\nabla \times E) \cdot da = -\frac{d}{dt}\Phi$$

Magnetic Flux through a surface:

$$\Phi = \int B \cdot da$$

???

TFM

4. Nov 13, 2008

### gabbagabbahey

Yes, now put the two equations together.

5. Nov 13, 2008

### TFM

Re: Faraday's Law and Stokes' Law

I had a feeling that was going to be the next stage...

so:

$$\oint_S (\nabla \times E) \cdot da = -\frac{d}{dt}\Phi$$

and

$$\Phi = \int B \cdot da$$

thus:

$$\oint_S (\nabla \times E) \cdot da = -\frac{d}{dt} (\int B \cdot da)$$

Since we hav the interagal of da on both sides, can this cancel down to:

$$\nabla \times E = -\frac{d}{dt} B$$

???

TFM

6. Nov 13, 2008

### TFM

I have a feeling that this is what we are looking for... it is one of Maxwells Equations...

7. Nov 13, 2008

### gabbagabbahey

Technically, the reasoning should go like this: (1) since the integral on the right is over space, we can tke the time derivative inside the integral so we get:

$$\int_S (\nabla \times \vec{E}) \cdot \vec{da} = \int_S -\frac{d}{dt} ( \vec{B} \cdot \vec{da}) = \int_S -\frac{d \vec{B}}{dt} \cdot \vec{da}$$

(2)Since this must hold true for any surface, the integrands must be equal...

$$\nabla \times \vec{E}= -\frac{d \vec{B}}{dt}$$

Saying "Since we hav the interagal of da on both sides, can this cancel down to:" is not a very good argument; it sounds like a 'hand-waving' argument to me and probably says to the person marking your homework that you don't really understand vector calculus.

You also may have noticed that I added in arrws to show explicitly which quantities are vectors and which are scalars....without arrows or boldface letters or something to distinguish the two, you demonstrate to the teacher that you're not sure which are which.

Also, the surface integrals don't need to be closed, they are over any surface; that's why I have written them as $$\int_S$$ instead of $$\oint_S$$....it's only the path integrals in Stokes' Law that need to be closed paths; since they are over the boundary of the surface, which is always a closed path even for an open surface.

Last edited: Nov 13, 2008
8. Nov 13, 2008

### TFM

Re: Faraday's Law and Stokes' Law

I see. That does make sense

Thanks,

TFM