Conceptual difficulty with faradays law

[DivergentSpectrum]

As i understand, the induced current in a wire equals -d(∫∫B⋅dS)/dt
What is S? I know it has to be an open surface for the flux integral to be non-zero, but over what surface do i do the integral?

 

[Doc Al]

As i understand, the induced current in a wire equals -d(∫∫B⋅dS)/dt

That should be the induced EMF, not current.

What is S? I know it has to be an open surface for the flux integral to be non-zero, but over what surface do i do the integral?

Consider the surface bounded by the wire.

 

[DivergentSpectrum]

Oh that's right the voltage not current (i don't know why they say emf is a force if its really a potential)
so the loop has to be closed then?

 

[jtbell]

The loop has to be closed. In principle, you can use any open surface whose boundary is the loop in question. In practice, you use a surface that makes it easy to calculate the flux of $\vec B$.

 

[DivergentSpectrum]

Thanks that makes since I was just thinkin in terms of some general wire but I guess the circuit would have to be closed for you to attach a voltometer.

So if the circuit is open there is no voltage generated?

 

[jtbell]

A changing magnetic field is associated with an electric field, according to Maxwell's equations: $$\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$$ or $$\oint {\vec E \cdot d \vec l} = - \frac{d}{dt} \int {\vec B \cdot d \vec a}$$ The electric field in turn produces an emf along a specified path: $$\mathcal{E} = \int {\vec E \cdot d \vec l}$$ So I would say the emf exists because the electric field does, regardless of whether there is a wire along that path or not, or a voltmeter to measure it. In order to produce a steady-state current, you need a circuit, i.e. a closed loop of wire that electrons can travel along.

 

[DivergentSpectrum]

A changing magnetic field is associated with an electric field, according to Maxwell's equations: $$\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$$ or $$\oint {\vec E \cdot d \vec l} = - \frac{d}{dt} \int {\vec B \cdot d \vec a}$$ The electric field in turn produces an emf along a specified path: $$\mathcal{E} = \int {\vec E \cdot d \vec l}$$ So I would say the emf exists because the electric field does, regardless of whether there is a wire along that path or not, or a voltmeter to measure it. In order to produce a steady-state current, you need a circuit, i.e. a closed loop of wire that electrons can travel along.

interesting... i wish the book i read had put it that way. i never even knew the curl of the electric field could be non-zero this really gives me a whole new perspective of what's going on here

 

[jtbell]

i never even knew the curl of the electric field could be non-zero

You started with electrostatics in which $\vec E$ does not change with time, and magnetostatics in which $\vec B$ does not change with time. In those situations, $\vec \nabla \times \vec E = 0$ so $\vec E$ is a conservative field and you can define the electric potential. Now you're starting on electrodynamics (time-varying $\vec E$ and $\vec B$ fields) in which this is no longer the case.
Pretty soon you'll see Maxwell's Equations all together in their glory.

(This stuff was what made me decide to be a physics major, by the way!)

 

[DivergentSpectrum]

lol awesome i know relativity is up next so I am getting pretty excited