Timeline for What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field?
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35 events
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| S Sep 30, 2021 at 6:08 | history | bounty ended | Brian | ||
| S Sep 30, 2021 at 6:08 | history | notice removed | Brian | ||
| Sep 26, 2021 at 6:37 | comment | added | alejnavab | “ The voltmeter is supposed to measure the voltage but there is no notion of voltage potential” // Hm, I’d say there is a voltage. It simply now depends on the path i.e. how the probes are placed in space. | |
| S Sep 26, 2021 at 5:48 | history | bounty started | Brian | ||
| S Sep 26, 2021 at 5:48 | history | notice added | Brian | Reward existing answer | |
| S Aug 31, 2021 at 9:25 | history | bounty ended | Brian | ||
| S Aug 31, 2021 at 9:25 | history | notice removed | Brian | ||
| S Aug 27, 2021 at 8:53 | history | bounty started | Brian | ||
| S Aug 27, 2021 at 8:53 | history | notice added | Brian | Reward existing answer | |
| Aug 27, 2021 at 8:45 | answer | added | Silicon Soup | timeline score: -1 | |
| Apr 5, 2021 at 23:35 | comment | added | wbeaty | @SredniVashtar ah true. dB/dt must stay inside! PS I wonder if people here encountered AJP papers like aapt.scitation.org/doi/abs/10.1119/1.12923 and aapt.scitation.org/doi/full/10.1119/1.1997171, plus my own semi-crackpot suggestion: make some nonconducting "wires" composed of mega-mu ferrite. Then wind them to form coils, to generate e-fields for electrostatic motors and solenoids. And, try some "narrow, very thick capacitor" wires made of PZT ferroelectric. Take all things we thought we understood, then "go perpendicular:" See my amasci.com/elect/mcoils.html [Edited] | |
| Apr 5, 2021 at 22:27 | comment | added | Sredni Vashtar | @wbeaty no more room in the answer and too long for a comment: i.postimg.cc/SQX7vYjj/rateofchange.png | |
| Apr 5, 2021 at 18:38 | vote | accept | Brian | ||
| Mar 25, 2021 at 23:54 | comment | added | wbeaty | Keep the voltmeter connected as shown, but move it to the other side of the schematic, and the reading will change. THE VOLTMETER CREATES A NEW 1-TURN COIL, and still gives a reading even when shorted. So instead, remove the voltmeter, and insert a current meter in series, which creates an oldschool voltmeter with Zinp of 3*R (add a new meter scale, calibrated in volts!) Also, "changing b-field" ...what, a NON-uniform one? Or, b-field caused by rf? If they didn't state "uniform b-field, in low-freq nearfield," then the voltmeter reading varies w/frequency and physical shape of the circuit. | |
| Mar 20, 2021 at 20:00 | history | edited | Brian | CC BY-SA 4.0 |
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| Mar 8, 2021 at 21:00 | history | tweeted | twitter.com/StackElectronix/status/1369030178731614210 | ||
| Mar 8, 2021 at 19:31 | history | edited | Brian | CC BY-SA 4.0 |
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| Mar 7, 2021 at 5:42 | comment | added | user57037 | Or one must give up on KVL in a Romer-Lewin type experiment. You can still do KCL, though. The current in the loop is the same throughout. AND you can even measure it if you want using the resistors as current sense resistors (as long as you pay attention to wiring details). | |
| Mar 7, 2021 at 4:23 | comment | added | Math Keeps Me Busy | @mkeith "The other problem is that as KVL is normally taught, you sum voltages based on nodes." However, nodes have absolutely nothing to do with KVL. KVL is about loop segments. "And KVL does say that the voltages in a loop sum to zero." One must account for the induced emf one way or another. Either include it as a lumped element, or include it as a term in the total emf around the loop. Could you demonstrate how you would calculate the displayed voltages in a Romer-Lewin type experiment? I can using the fact that the total emf in a loop equals total voltage drop. Can you without that? | |
| Mar 7, 2021 at 3:49 | comment | added | user57037 | The other problem is that as KVL is normally taught, you sum voltages based on nodes. But in this circuit it is not enough to specify the node. I don't know. I think you are stretching it. And KVL does say that the voltages in a loop sum to zero. That is KVL. @MathKeepsMeBusy | |
| Mar 6, 2021 at 23:37 | comment | added | Math Keeps Me Busy | @mkeith I agree that, in the case of time varying magnetic fields, there is no longer a uniquely defined potential difference between two points. You agree that "if you know the path, you can calculate...". But KVL applies to specified paths, not to arbitrary points. "And the loop doesn't have zero potential, either, which is the essence of KVL." The essence of KVL is that the total emf of a loop equals the total voltage drops across the segments of the loop. That remains true. I have updated my answer, please take a look. | |
| Mar 6, 2021 at 22:28 | comment | added | user57037 | @MathKeepsMeBusy the potential between two points is no longer uniquely defined. I don't really see KVL being salvageable. But if you know the path you can calculate the voltage. That much is true. It is just that there are other paths and they will give you different voltages. So the concept of nodes in circuit analysis is gone right out the window. And the loop doesn't have zero potential, either, which is the essence of KVL. | |
| Mar 6, 2021 at 22:14 | comment | added | Math Keeps Me Busy | @mkeith I disagree. If time varying flux lines cut your circuit, then "potential differences" are out. However, KVL, if understood as follows, still applies. My understand of KVL. The sum of all the voltage drops along a loop equals the total emf around the loop. Voltage drop being defined as the work done by an electron to move from one place to another along a given path. Since the path is defined for any loop, voltage drop is defined for one point to another along a curve, even though a curve-independent potential is not. | |
| Mar 6, 2021 at 8:00 | comment | added | user57037 | The engineer/technician attitude is that the "true" voltage must be the voltage across one of the resistors. Any voltage induced in the leads of your measurement apparatus is dismissed as "measurement error." So I would tell you to place the leads directly at the resistor, and twist the wire as it moves away from the resistor all the way back to the voltmeter to cancel out any inductive pick-up. If you do this, you will get V = I * R right at the resistor. But Lewin is NOT WRONG. He is a smart and well educated guy. Electroboom is not wrong either. If flux lines cut your circuit, KVL is out. | |
| Mar 6, 2021 at 5:17 | answer | added | Kinka-Byo | timeline score: 3 | |
| Mar 5, 2021 at 13:44 | answer | added | Sredni Vashtar | timeline score: 15 | |
| Mar 5, 2021 at 11:29 | answer | added | user215805 | timeline score: 2 | |
| Mar 5, 2021 at 11:21 | history | edited | Brian | CC BY-SA 4.0 |
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| Mar 5, 2021 at 9:58 | history | edited | Brian | CC BY-SA 4.0 |
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| Mar 4, 2021 at 13:16 | comment | added | Brian | Ohh good point @rdtsc, I've edited that out now. | |
| Mar 4, 2021 at 13:16 | history | edited | Brian | CC BY-SA 4.0 |
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| Mar 4, 2021 at 13:11 | comment | added | rdtsc | The voltmeter is measuring the current created by a moving magnetic field; a stationary field generates zero current. | |
| Mar 4, 2021 at 13:11 | answer | added | Math Keeps Me Busy | timeline score: 4 | |
| Mar 4, 2021 at 13:09 | history | edited | JRE | CC BY-SA 4.0 |
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| Mar 4, 2021 at 12:40 | history | asked | Brian | CC BY-SA 4.0 |
