Is this true about the emf in a closed circular wire?

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SUMMARY

The discussion centers on the concept of electromotive force (emf) in a closed circular wire, specifically addressing the relationship between potential difference and the presence of a battery. It is established that when traversing a closed loop containing a battery, the potential difference is not zero due to the battery's contribution to the circuit. The key point is that the line integral of the electric field around a closed path is not zero because the circuit includes an emf source, which introduces a discontinuity in the potential difference.

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latentcorpse
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the emf is defined as the potential difference between two points \varphi(\vec{r_1})-\varphi(\vec{r_2}).

ok so let's say we make a trip round a closed circular wire with a battery to keep the current flowing then r1=r2 and so no emf has been done - is this true and if so why?
 
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The battery in general produces an emf. r1 does not equal r2 because you hit the battery on your walk around the loop. If r1=r2 you can't fit a battery in there.

I'm guessing you're talking about Kirchoff's rules? If you're talking about why the line integral of the Electric field around a closed path is not identically 0, it's because currents are not static.
 


i don't follow. i haven't covered kirchhoffs rules yet. why does r1 not equal r2, these are just vectors from the origin to the point we're at in the circuit so surely regardless of whether we do a lap or not its still the same position vector from the origin that descirbes the point?
 


When you walk around the loop, you will inevitably hit the battery or emf source. That source is like a discontinuity in your loop...when you "walk" over the emf source you need to add the voltage given off by the emf to your potential difference.
 

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