Is an induced emf always produced by a change in flux?

[James Bond]

Faradays law tells us that a change in flux induces an emf. Now consider the phenomenon of motional emf. It is observed across the ends of an open conductor (ie one which is not in a circuit). It is always discussed in connection to faradays law. Where is the change in flux in the case of a straight conductor moving perpendicular to a uniform magnetic field?
Also, is the converse of faradays law true? That is, is an induced emf ALWAYS produced by a change in magnetic flux?

 

[BvU]

Welcome to PF, commander :)

The underlying "mechanism" is the Lorentz force, for which see google. That brought about the Maxwell[/PLAIN] equations, which see idem. Faraday is basically one of them. (But the guy didn't know that at the time: Maxwell was born just about then) .

 

[James Bond]

Thanks for the greetings, BvU!
So I guess the only interpretation of faradays law is through the lorentz force interpretation. Thanks for your help!

 

[Philip Wood]

I used to think that motional emf could be subsumed into a rate-of-change-of-flux-linkage phenomenon by incorporating the moving conductor into a closed loop (which need not even be conducting). Then the loop area changes as the conductor moves, so there's a change of flux linkage. But there are cases like the Faraday disc which it's difficult to see as cases of circuit area changing. I believe Feynman cites other such cases. So I think that for a moving conductor it's better to regard \frac{d \Phi}{dt} as rate of cutting of flux, which it's easy to show is equivalent to (\vec{v}\times \vec{b}).d \vec{l}, based on the magnetic Lorentz force.

 

[vanhees71]

It's good to keep in mind that the fundamental physical laws underlying classical electromagnetics are the local Maxwell equations, i.e., the Maxwell equations in differential form. Much of the problems students have with the Law of Induction is that it's often represented in incomplete form. The fundamental Faraday law reads
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
Using Stokes's integral Law for an arbitrary surface $S$ with boundary $\partial S$, gives
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_S \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now comes the tricky business. To get the usual Faraday Law in integral form, you want to take out the time derivative from the surface integral on the right-hand side. Often people don't discuss this carefully enough. If you have a moving surface, there's an additional term. Taking this properly into account, you'll get the one and only correct integral form of Faraday's Law of Induction
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
Here $\vec{v}=\vec{v}(t,\vec{x})$ is the velocity field of the boundary of the surface. This tells you that you have to define the electromotive force in Faraday's law including the complete Lorentz force per unit charge and not only the electric piece. Of course, if the surface under consideration is not moving, then $\vec{v}=0$, but only then it's correct to forget the magnetic part of the Lorentz force!