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

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?

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BvU
Homework Helper
Welcome to PF, commander :)

The underlying "mechanism" is the Lorentz force, for which see google. That brought about the Maxwell[/PLAIN] [Broken] 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) .

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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
Gold Member
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
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_S \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$