# Inducing a current

by jsmith613
Tags: current, inducing
 Sci Advisor PF Gold P: 11,948 'A scientist' says that there may be much more water in the earth's mantel than in all the Oceans.
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P: 3,443
 Quote by cabraham That is my point. Maybe I'm belaboring something not quite within the OP intentions. But several posters stated that an emf exists in free space. I felt compelled to point out that it is a very ambiguous ill-defined non-unique value of emf. In order to have any meaning to the emf concept, a conductor mast be placed in the region and the shape of said conductor determines the emf. Rearranging said conductor shape changes the emf. That is all I wish to emphasize. Sorry if I got carried away. BR.
That's alright. jsmith said he/she? would be away until Sunday anyway. You're right, the closed integral of E*dL does depend on the path. But this doesn't mean it is ill-defined, it just means we must specify the path. If we define the integral of E*dL as the emf, then the emf depends on the path specified (in other words, asking about the emf between two points is useless, unless we also state what path it is taken over). There is no problem with this. It is the more correct way of thinking. When we start talking about a wire, then the path is implied to be through the wire, but if we were speaking more accurately, we would also need to state explicitly what the path is.

I think in an earlier post you said something like 'what use is emf when there are no charges or currents?' And this is a good question. I can't think of any use in classical EM. And I doubt there is any use in EM without charges or currents in general relativity (although I'm not sure, because I don't know a lot about general relativity). This is probably why a lot of people define emf as existing only in a circuit - because there is not much use for the concept when there is no circuits. I would not impose this extra restriction, because I don't like to complicate a definition any more than is necessary, but it just depends on personal preference.
 HW Helper P: 3,443 About this idea of a displacement current which causes a current to flow despite a very large resistance of the oscilloscope: This would mean we are saying the scope effectively acts as a capacitor, right? At first guess, I would have thought that the scope is designed so that it does not act as a capacitor. Also, there is the issue that the induced electric field might not change quickly enough to allow a significant displacement current. Both these issues are difficult to calculate. I Looked it up on wikipedia, found: "General-purpose oscilloscopes usually present an input impedance of 1 megohm in parallel with a small but known capacitance such as 20 picofarads." So our equation for the current is: $$I = N C \frac{\partial^2 \Phi}{\partial t^2}$$ (where N is the number of turns of the coil, and C is 20 picofarads). So our answer depends on the twice differential (with respect to time) of the magnetic flux. This is a bit more difficult to estimate. We can say we are using a magnetic dipole, which is falling under gravity, and the magnetic flux will be the integral of the magnetic field over the area inside the coil (which will vary at different points in the area, and depend on the distance of the area from the dipole). Difficult :( Anyway, I should go to bed. I should have been asleep already!
P: 614
 Quote by BruceW When the magnet is in the middle of the coil, you are right that there is zero emf across the entire coil. When the magnet is above the middle of the coil, the current will flow anticlockwise. This movement of charge will decrease the rate of change of flux below the magnet, but will increase the rate of change of flux above the magnet. But because the magnet is nearer the top half of the coil, the effect of the current flowing above the magnet is less than the effect of the current flowing below the magnet, so overall the rate of change of flux is decreased. From this explanation, you can see that the braking effect will be zero when the magnet is in the middle of the coil, since there is as much coil above as there is below, so if there was a current, it would have no effect on the magnet's motion.
back a little early!!

(from the oscilliscope we can see the emf is zero for the entire time the magnet is in the coil)

so lets say we connected an ammeter instead of an oscilliscope as a current would definitly flow then. you seem to be saying we would only get no breaking effect when the magnet is EXACTLY in the centre of the long coil.
using your logic (when the magnet is above the centre there is a greater change in flux beneath the magnet) there should be an emf.
I think a better way to think of it is the amount of flux linking with the coil. When the entire magnet is inside the coil there is no change of flux linking with the coil so no emf is induced.

Having never performed the experiment with a coil I actually don't know the results but I would presumed, base on my way of thinking we would only get magnetic breaking at the ends of the magnet (i.e: as it enters and as it leaves).

I may be getting terribly confused in which case I apologise.
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 Quote by jsmith613 I may be getting terribly confused in which case I apologise.
No, you're not confused, you're pretty much right. But the emf would be slightly non-zero when the magnet was in the coil. If you think about it, the graph of emf is smooth, this is because there is still a slight emf even when the magnet has entered the coil.

The other way to think about it, is that when the magnet is inside the coil, there is less flux linkage than if the magnet was right in the middle of the coil, so the flux linkage is still changing even when the magnet is inside.

But I think you are right, that the emf when the magnet is inside the coil is much less than when the magnet is entering/leaving the coil. (Especially if the coil was quite long, the emf would be very small when the magnet was inside).

So for an exam, I would just say there is an emf when the magnet is entering/leaving. (Since the emf when the magnet is inside will be an order of magnitude smaller). I hope I haven't made things unnecessarily confusing.
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 Quote by BruceW No, you're not confused, you're pretty much right. But the emf would be slightly non-zero when the magnet was in the coil. If you think about it, the graph of emf is smooth, this is because there is still a slight emf even when the magnet has entered the coil.
The emboldened statement does not make sense but the rest of it does

so there will be no magnetic breaking effect (for A-level purposes), right?
thanks :)
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 Quote by jsmith613 so there will be no magnetic breaking effect (for A-level purposes), right? thanks :)
That's definitely the right A level answer for the coil.
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 Quote by sophiecentaur That's definitely the right A level answer for the coil.
cheers
once again I must express my gratitude to EVERYONE who has helped me in this discussion
I do hope that the tangent discussion interested other people as well....you can now discuss to your hearts content :)
 Sci Advisor PF Gold P: 11,948 No problem. The one big risk on this forum is over-answering questions. I'm sure it has given many A level students real headaches. "I only wanted to know . . . . . ." Even when the questioner states his or her level of interest, that doesn't necessarily deter the enthusiastic contributor. Good luck with the A level stuff.
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 Quote by jsmith613 The emboldened statement does not make sense but the rest of it does so there will be no magnetic breaking effect (for A-level purposes), right? thanks :)
For A-level purpose, yes, there is only a braking effect when the magnet enters and leaves the coil. I think you've got a pretty good grasp of this stuff. Good luck anyway, even though I don't think you'll need it :)

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