"Here, all arcs are isolated wires, having a voltage means induction in open circuit. If we keep the idea of non induction in open wire, then the voltage in the arcs should decrease progressively to zero when the resistance increases to infinity, that is, the induced voltage should depend on current."
Why do you think that voltage does not get induced in the wire when it is broken? The voltage that gets induced is induced in the region where the magnetic field is changing - it is independent of what is in that region (not counting if the region is a conductor, gets current, and in turns, affects the magnetic field and induced voltage in the area). How would a voltage be measured across each resistor? - by attaching a voltmeter in parallel to each resistor. Typically a voltmeter acts as a huge resistor so that almost no current goes through it. As such it does not affect the circuit. More so, because it is in parallel, the resistor you attach it to cannot affect the voltage measured by the meter. So keeping the situation the same, and increasing the resistance to infinity would not change the voltage measures. Cutting a wire is like increasing the resistance to a very very large number (air gap). You can increase it further with insulators, but at no point will it be actually infinite. So, even if your argument was correct, since an open circuit and a circuit with huge resistors is electrically identical, the behavior will not be different compared to using smaller resistors.
I pose two possibilities: 1) there is induction in open circuit, this is not allowed by Faraday's law; 2) There is not induction in open circuit, then Faraday's law is wrong because the voltage must vary.
This is a paradox that Faraday's law cannot solve.
Faraday's law has to do with wires, but the generalized form "Maxwell–Faraday equation" http://en.wikipedia.org/wiki/Faraday's_law_of_induction has to do with the region. The electric field is in the vicinity of a changing magnetic field as per that equation. If you put a charged particle in that field, it will accelerate accordingly. If you place a conductive wire in that area, the charges will move accordingly. If you were to place a rod of conductive metal (open circuit), the electrons would move to one side until force on the charges in the wire from the field produced by the separation of charges equals the force on the charges due to the induced field in space.
In fact, I was battling with someone who claimed firmly that Faraday's law is valid only for closed circuit. This is why I said "there is induction in open circuit, this is not allowed by Faraday's law". For "Maxwell–Faraday equation", it cannot stand because in a loop, the line integral of E is zero inside the wire. If you joint the ends of the wire together without making contact, the closed line integral of E is zero, while that of "Maxwell–Faraday equation" must equal -d phi / dt. This is the sense of my Faraday's law paradox explained in these articles: http://pengkuanem.blogspot.com/2015/02/coil-and-resistor-induction-paradox.html http://pengkuanem.blogspot.com/2012/10/faradays-law-paradox.html
Otherwise, in reality, there is induction in antenna which are open circuits.
"Here, all arcs are isolated wires,
ReplyDeletehaving a voltage means induction in open circuit. If we keep the
idea of non induction in open wire, then the voltage in the arcs
should decrease progressively to zero when the resistance
increases to infinity, that is, the induced voltage should depend on
current."
Why do you think that voltage does not get induced in the wire when it is broken? The voltage that gets induced is induced in the region where the magnetic field is changing - it is independent of what is in that region (not counting if the region is a conductor, gets current, and in turns, affects the magnetic field and induced voltage in the area).
How would a voltage be measured across each resistor? - by attaching a voltmeter in parallel to each resistor. Typically a voltmeter acts as a huge resistor so that almost no current goes through it. As such it does not affect the circuit. More so, because it is in parallel, the resistor you attach it to cannot affect the voltage measured by the meter. So keeping the situation the same, and increasing the resistance to infinity would not change the voltage measures. Cutting a wire is like increasing the resistance to a very very large number (air gap). You can increase it further with insulators, but at no point will it be actually infinite. So, even if your argument was correct, since an open circuit and a circuit with huge resistors is electrically identical, the behavior will not be different compared to using smaller resistors.
I pose two possibilities: 1) there is induction in open circuit, this is not allowed by Faraday's law; 2) There is not induction in open circuit, then Faraday's law is wrong because the voltage must vary.
DeleteThis is a paradox that Faraday's law cannot solve.
Faraday's law has to do with wires, but the generalized form "Maxwell–Faraday equation"
Deletehttp://en.wikipedia.org/wiki/Faraday's_law_of_induction
has to do with the region. The electric field is in the vicinity of a changing magnetic field as per that equation. If you put a charged particle in that field, it will accelerate accordingly. If you place a conductive wire in that area, the charges will move accordingly. If you were to place a rod of conductive metal (open circuit), the electrons would move to one side until force on the charges in the wire from the field produced by the separation of charges equals the force on the charges due to the induced field in space.
In fact, I was battling with someone who claimed firmly that Faraday's law is valid only for closed circuit. This is why I said "there is induction in open circuit, this is not allowed by Faraday's law". For "Maxwell–Faraday equation", it cannot stand because in a loop, the line integral of E is zero inside the wire. If you joint the ends of the wire together without making contact, the closed line integral of E is zero, while that of "Maxwell–Faraday equation" must equal -d phi / dt. This is the sense of my Faraday's law paradox explained in these articles:
Deletehttp://pengkuanem.blogspot.com/2015/02/coil-and-resistor-induction-paradox.html
http://pengkuanem.blogspot.com/2012/10/faradays-law-paradox.html
Otherwise, in reality, there is induction in antenna which are open circuits.