Let us say that there is an electrical circuit and it also has a inductor of inductance L. The current on the circuit is a function of time, I(t). So now we can be sure that there is always a potential drop across the inductor. And let us assume that dI(t)/dt = a, so potential difference across the inductor is L x a.
So, here's my question! Will a Voltmeter connected across the inductor show the same value of potential difference, i.e. L x a ?
(Also, Potential difference holds no meaning here as the Electric Field induced is non-conservative!)
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thats a very good question
and i think the answer.. will totally depend on the orientation of the voltmeter.. U should watch
although its not exactly what you asked.. the concept is the same.. calculating "potential difference" (which really makes no sense when we have induced fields) between 2 points by using a voltmeter..
and two same voltemters give different readings.. based on their orientation.. Great/Legendary Professor Walter lEwin demonstrates this spectacular phenomena.. which for me is trully amazing!!!
I have seen the whole lecture by the way! But this question is still unanswered! :(
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Because voltage across the inductor is: Ldi/dt and di/dt=d(at)/dt=a, then V=La
This is not as simple as you feel it is! Voltage Difference golds no meaning when you talk about non conservative electric field. You may easily solve a problem by using Kirrchoff rules here but that may be mathematically correct but is actually wrong. It "might be" correct i.e., L x a may be the answer, but a proper explanation would help!
If we have to hold a discussion about inductors behavior based on your understanding of the relation V=d(Li)/dt as a consequence of Kirchoff´s Laws, then I give up here.
I just want a proper explanation, buddy! Tell me if I went wrong somewhere! Thats it.
The law that relates induced voltage and current in an inductor can be found in Maxwell´s equations (3rd equation in differential form) or in Farady's Law
I know that. Let's not be mathematical, why is it, that's my question!
It's because in the case of the inductor, the wire connecting the voltmeter to the system does not go INSIDE the coil, but is always outside. Th coil has a symmetry of revolution about its axis, so, whatever the place the voltmeter is, it will always read the same.
yup.. that is a good enough explanation i feel :) :) by mister Vincent!
@Vincent-Lyon.Fr. Some contradiction in your theory? If there is symmetry outside, there is also inside then the fact that the wire goes inside the coil or not is meaningless.
i think by symmetry what he means is that, all the path that the current can choose to flow through will turn out to be the same, and thus the "potential difference" will have the same value!
True, if the connecting wire goes inside, the revolution symmetry will give the same reading on the voltmeter, whatever the orientation, BUT it will be different from the one measured outside.
The one measured outside is not affected by the changing B-field and induced emf of the coil.