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$$\nabla \times \vec{E} =-\frac{\partial{\vec{B}}}{\partial{t}}$$

Applying Stokes' theorem:

$$\oint_{loop} \vec{E} \cdot d\vec{l}=\int_S -\frac{\partial{\vec{B}}}{\partial{t}} \cdot d\vec{S}$$

Where $S$ is any open surface with the loop as its perimeter. Considering the simplicity of the derivation I struggle to see why this shouldn't hold in all cases. However, consider the setup described below:

A uniform magnetic field parallel to the $z$ axis crosses the $x$-$y$ plane in a circular region radius $r$ centred at the origin. A wire loop of radius $R>r$ is also centred at the origin in the $x$-$y$ plane. The field changes with time (but remains parallel to the $z$ axis). Is an EMF induced in the loop?

According to Faraday's law as expressed above, the answer surely should be yes. The integral on the right hand side will not return 0, so therefore the left hand side cannot be 0 and an EMF should be induced in the coil.

Physically, however, the answer surely should be no. An EMF is induced in the wire ultimately because of the magnetic force that acts on the charges in the loop itself, moving them along the loop and producing a potential difference. If the loop itself is not in the magnetic field, then how can the field exert a force on the charges in it?

There is an error in reasoning somewhere here. Where am I going wrong?

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marked as duplicate by ZeroTheHero, John Rennie electromagnetism Apr 5 at 8:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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(a) It's not the magnetic field that exerts forces on the charges in the loop, but the electric field given by your second equation. [For one thing the charges in the wire are (on average) stationary, so wouldn't experience forces from a magnetic field even if it were local to the charges.]

(b) The line integral of the electric field around the loop gives the emf in the loop, but the concept of potential difference doesn't apply because the field is not conservative: if we take two points A and B on the loop, the work done by a charge going from A to B depends on the route by which it goes from A to B.

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  • $\begingroup$ Thanks for pointing out that the electric field in this case isn't conservative, although frankly that's blown my mind. We use the idea of potential difference all the time in circuit theory, and indeed that's what EMF is (an induced potential difference). Clearly though this can't be potential difference the same way we describe it in electrostatics. Is potential difference in circuit theory just loose terminology? $\endgroup$ – Pancake_Senpai Apr 4 at 17:06
  • $\begingroup$ It's fine to use the concept of pd when we're outside the seat of an emf, for example in the circuitry external to a battery or generator. Worry not. There are many good posts about emf and pd on this site. $\endgroup$ – Philip Wood Apr 4 at 17:54
  • $\begingroup$ @Pancake_Senpai EMF is not the same thing as potential difference. For a stationary wire near changing magnetic field, EMF is integral of electric field strength along this wire, but potential difference is integral of electrostatic part of this field, along any path that connects the two ends. $\endgroup$ – Ján Lalinský May 20 at 22:19

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