# Conductors and induced emfs: an inconsistency?

I quote from Principles of Electricity by Page and Adams (P and A), an elementary textbook, well-respected in its day.

"[A]lthough an emf acts round a fixed curve through which the [magnetic flux] is changing, whether a conducting circuit coincides with the curve or not, we cannot suppose a motional emf to exist in the absence of moving charges, such as the electrons carried along by a moving conductor." [My italics]

The first sort of electromagnetic induction mentioned above is described by the integrated form of the Faraday-Maxwell equation $$\mathscr E=\int_{\partial S} \mathbf E.dl=-\frac{d}{dt}\int_S\mathbf B.d\mathbf S$$ I've always accepted that this applies, as P and A claim, to any stationary closed loop, whether or not it conducts. But one notes that, these days, $$\mathbf E$$ is usually defined in terms of the electric Lorentz force, $$q \mathbf E$$, on some test charge, $$q$$. And when explaining the significance of $$\mathscr E=\int_{\partial S} \mathbf E.dl$$ it is very tempting to introduce at least one charge that can go round the loop and have work done on it.

But if we're prepared to accept that the stationary loop needn't be a conductor , why can't we say the same for a loop parts of which are moving in a non-time-varying magnetic field? A "raw" form of the equation for the emf in such a case is $$\mathscr E=\int_{\partial S}(\mathbf v \times \mathbf B).d \mathbf l$$ in which the element $$d\mathbf l$$ has velocity $$\mathbf v$$ and is in a local magnetic field $$\mathbf B$$.

The answer that springs to mind immediately is: $$(\mathbf v \times \mathbf B)$$ is the magnetic Lorentz force per unit charge, and you can't have such a force without a moving charge. But then, going back to the case of changing magnetic flux through a stationary circuit, we seem to accept the integration of $$\mathbf E$$ around the loop, without worrying that $$\mathbf E$$ is the electric Lorentz force per unit charge and insisting that charges be present!

On the face of it there is an inconsistency. I'd like someone to explain the error(s) in my reasoning.

• $\bf E$ is defined in terms of the work that would be done on a hypothetical test charge, were it to be placed in the field and moved around. But the test charge doesn't have to be placed there for the field to exist. If there's some reputable source that says it does, please provide a link or reference. Commented Jan 17, 2020 at 22:31
• Agreed (though I think you mean $\mathscr E$, that is emf, rather than $\mathbf E$). My question is: why can't we say the same for motional emf, substituting $\mathbf v \times \mathbf B$ for $\mathbf E$, and not insisting on a physical conductor. Commented Jan 17, 2020 at 22:39
• Sorry, I did mean $\bf E$, but I should have said it's defined in terms of the force on a hypothetical charge, not the work done. Commented Jan 17, 2020 at 22:41
• Possible duplicate where I and others address this issue: physics.stackexchange.com/questions/525372/… Commented Jan 17, 2020 at 22:44
• "who ever did say that the charge must actually exist to define a motional emf?" That was in the Page and Adams quote that I gave at the beginning. P and A, though first year university level only, was rather good on basic concepts. But I suspect that here they were wrong. And it looks as if you think that, too! Commented Jan 17, 2020 at 23:38