Induced current of this conductor?

  • #1
PhiowPhi
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I have a confusing example that I'm studying, I tried to figure out the direction of induced current. But the right hand rule would tell me there isn't, but there has got to be since there is change in flux?!

Here is the example:

vO4tYAD.png


The gray bar is the conductor that is moving through a magnetic field with velocity(v marked), the orange wires are connected to circuit outside the diagram, the polarization is what I think the induced EMF is going to be, when using the right hand rule the induced current is pointing up. I'm certain there is induced current, but not sure where it will flow.

In contrast to this diagram:

FraGB9L.png


It's relatively easy to figure things out.
 

Answers and Replies

  • #2
zoki85
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I don't know what is confusing you. Induced emf is E = (v×B)⋅l .
In first case there's no current flowing in the outside circuit, and in the second case there could be if the circuit is closed.
 
  • #3
cabraham
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I don't know what is confusing you. Induced emf is E = (v×B)⋅l .
In first case there's no current flowing in the outside circuit, and in the second case there could be if the circuit is closed.

For an antenna, there is current even though the circuit is not closed. At high enough frequency the displacement current can be substantial. Current can and does exist in open circuits. The higher the frequency, the greater the current.

Claude
 
  • #4
zoki85
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For an antenna, there is current even though the circuit is not closed. At high enough frequency the displacement current can be substantial. Current can and does exist in open circuits. The higher the frequency, the greater the current.

Claude
Let's not complicate things when OP has difficulty with B=const, v=const. case.
 
  • #5
PhiowPhi
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I don't know what is confusing you. Induced emf is E = (v×B)⋅l .
In first case there's no current flowing in the outside circuit, and in the second case there could be if the circuit is closed.
Assuming the circuit is closed. Why wouldn't current flowing in the first? Due to the induced EMF being across the wire?
 
  • #6
PhiowPhi
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I guess it's because of the wires themselves in that diagram(1), wouldn't allow any current flow(if it we're a closed circuit), but if it we're like so:

tyo8wFT.png

It would?
 
  • #7
zoki85
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tyo8wFT.png

It would?
It could now . Note that in this case vertical length of wire contributes more volts to the properly closed circuit than length of the bar.
 
  • #8
PhiowPhi
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It could now . Note that in this case vertical length of wire contributes more volts to the properly closed circuit than length of the bar.

True, due to its L > than that of the bar.
However, what if that wire was not inducing any EMF of it's own(being fixed, or not in the magnetic field, etc...) would current still flow from the bar in a closed circuit?
 
  • #9
zoki85
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True, due to its L > than that of the bar.
However, what if that wire was not inducing any EMF of it's own(being fixed, or not in the magnetic field, etc...) would current still flow from the bar in a closed circuit?
Since the bar has some vertical length , some small emf would be induced in it and some small current would flow to external circuit.
 
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  • #10
PhiowPhi
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Since the bar has some vertical length , some small emf would be induced in it and some small current would flow to external circuit.

Makes sense, I guess as long as there is a path for current to flow on that wire(which it does in comparison to it's original diagram), it would induced EMF equal to it's short Length x(vB). And if I accounted the Length of those "additional" wires on the edges it would bump the induced EMF up, Thanks!
 
  • #11
PhiowPhi
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@zoki85 , I'm struggling a bit with the theory.
Going back to the 1st diagram, the reason why current cannot flow(assuming it's a closed circuit), it's because the force on the charge is upward and there is no path for it to move upward? And by attaching those orange wires to the upper/lower surfaces(ignoring their induced EMF's, just focusing on the gray bar's induced EMF) it creates a path for current to flow from high potential to low potential? - Just making sure I got the idea right.


Also, is width something to considering at all? I won't because E = -vBL, but... maybe I'm missing something.
 
  • #12
jim hardy
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I find it useful to imagine myself very small and sitting inside the conductor holding one of the free charges in my hand.
The free charge is an electron but QV cross B is written for positive Q, so we defer to convention...
Like you, in first picture i get Lorentz force on the positive charge is UP
which is same as Hall Effect were there current.
Now close the loop and you get force UP in both the orange wires, again no current flows because of equal and opposing voltages in the orange wires
that is, until your loop starts to exit the magnetic field as it travels right.
Then the induced voltage in right hand orange wire decreases because B out there has dropped to zero,
and as your loop moves right the flux it encloses is decreasing

So both formulas QV cross B and d(phi)/dt both predict a voltage in the loop and current if it's closed.

It is a valid logical tool to reduce things to this preposterous level of simplicity .
When your intuitive reasoning leads you to the same conclusion as the formulas you've memorized you are a whole lot better equipped for exams.
Strive for that 'gut feel' in your equations. Continually rework your mental images until they take you straight to the math. It's easier to figure out equations from good solid basics than to memorize them.

old jim
 
  • #14
PhiowPhi
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Thank you for the elaborative explanation @jim hardy , starting to get it(I hope), specially when imaging myself there. Lately I don't trust my gut feelings as much because it's wrong... and therefore need to review!

What's interesting though, is the following result when current "starts" to flow, if the conductor is connected to a circuit:
3PIXcET.png

Where those circular things are charges, and the yellow line represent the flow of this charge(i/e flow of current).

Is this correct? Based on where the wires are connected, and that's the only path the right most charge could move, and the middle and the left most charge just has to go up.


@zoki85 , thanks for the link, it's a great refresher. But even when I understood the theory, it's a bit confusing applying it so some examples, like this one.
 
  • #15
PhiowPhi
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Also, another way I'm imagining this is to think of EMF as pressure and current as water. An old analogy, but really useful when learning about such systems. It's also a reason it makes sense(from that perspective) as to why the first diagram, wouldn't allow any current flow. Due to the pressure being up, and there is no path for water to move up(somewhat).
 
  • #16
jim hardy
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Lately I don't trust my gut feelings as much because it's wrong... and therefore need to review!
That's good. You have to work it in your head until your intuition takes you to the formulas. When your intuition and the math agree you are getting someplace.
The mind can believe most anything so we must lead it to truth by checking and cross-checking our logic.

Is this correct? Based on where the wires are connected, and that's the only path the right most charge could move, and the middle and the left most charge just has to go up.
Your new drawing has both yellow end wires oriented so that the force on charges is now upward. A current could flow if the loop is completed outside the magnetic field where its vertical part doesn't induce an opposing voltage..

Also, another way I'm imagining this is to think of EMF as pressure and current as water. An old analogy, but really useful

Water analogies can be very useful. But one must be wary of a couple thought traps they can lead you to:

We pump water out of the ground, and water that's not contained will fall to the ground and seep in. This causes people to mistakenly think that "ground" has some magical attraction for electric current. In reality "ground" is just another wire that goes most everywhere.

We use a garden hose to transport water molecules from point A to point B. If your garden hose is full there's very little delay between opening the spigot and getting water out the far end. This causes people to mistakenly think that electrons move down a wire at near the speed of light, which they don't. They drift along slower than an ant's pace. A voltage change however propagates down the wire near speed of light, but that's just the charges bumping into one another.
A better analogy is a hydraulic power system.
A hydraulic power system is used to transport energy from point A to point B. Fluid is pumped into a hose at high pressure, and at far end of hose fluid is forced out by the fluid behind it to do work. The fluid moves slowly along the hydraulic hose, but a pressure change propagates down it at speed of sound.
When you push a few molecules of fluid in one end of the hose a similar number of molecules pop out the far end but they're not the same molecules .
When you push an electron's worth of charge into one end of a wire a similar electron's worth will pop out the far end, but it's not the same electron.

Our good friend SophieCentaur caused me to rethink the water analogy and I've added the above caveats to my use of it. He comes down on the concept of "electrons whizzing around a circuit at light speed", and rightly so..

In electric circuits we're usually more interested in transporting energy than in transporting charge.
So be rigorous in your use of water analogy and keep Pascal's principle in mind--You transmit energy through the fluid.
http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html

Have fun and good luck in your studies. old jim
 
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  • #17
PhiowPhi
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Thanks Jim! And zoki.
 
  • #18
cabraham
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Let's not complicate things when OP has difficulty with B=const, v=const. case.
To set the record straight is NOT "complicating things". Every person, even a beginner, should be given the truth from day one. It is more difficult this way, but it's better that the newbie not develop misconceptions since they are hard to correct later. I simply stated the truth that current does not require a closed path. What do you suggest telling him? That current does require a closed path? Then at what point do you suggest we correct the falsity? Just wondering.

Claude
 
  • #19
PhiowPhi
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Hey Claude, I appreciate it. But I think what's meant is you don't have to explain everything at once from the beginning, but gradually. I can't learn everything in one question, but then again... it's true to be given a few facts here and there in the beginning to set one on the right track.

Let a man crawl first, then let him start to walk, and then watch him run. Step by step.
 
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  • #20
cabraham
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Hey Claude, I appreciate it. But I think what's meant is you don't have to explain everything at once from the beginning, but gradually. I can't learn everything in one question, but then again... it's true to be given a few facts here and there in the beginning to set one on the right track.

Let a man crawl first, then let him start to walk, and then watch him run. Step by step.

I can agree to that (bold print). I just didn't want to let him labor for years under a false assumption. But your point is valid.

Claude
 
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  • #21
PhiowPhi
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Thanks Claude.
 
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