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epsilonjon
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Hi.
I have just started learning about inductors, and this is the method my book uses to show that Kirchhoff's second law is still valid even with inductors in a circuit, and to calculate the voltage drop across an inductor:
"According to Kirchhoff's loop rule, the algebraic sum of the potential differences around any closed circuit must be zero because the electric field produced by charges distributed around the circuit is conservative. We denote this such a conservative field as \vec{E_{c}}.
When an inductor is included in the circuit, the situation changes. The magnetically induced electric field within the coils of the inductor is not conservative. We denote this field as \vec{E_{n}}. We need to think very carefully about the roles of the various fields. Let's assume we are dealing with an inductor whose coils have negligible resistance. Then a hegligibly small electric field is requiresd to make charge move through the coils, so the total electric field \vec{E_{c}} + \vec{E_{n}} within the coils must be zero, even though neither field is individually zero. Because \vec{E_{c}} is nonzero, we know there have to be accumulations of charge on the terminals of the inductor and the surfaces of its conductors, to produce this field."
The author then goes on to use this, together with Faraday's law and the self-induced emf of the inductor, to work out the voltage drop across it. He concludes that we are fine in using Kirchhoff's second law, so long as we are defining the voltages in terms of the conservative part of the field.
I am trying to satisfy myself that this is true, but I have a problem: what if I apply the same thinking (regarding the total electric field being zero) to just a circular loop of wire in a changing magnetic field?
I know that an electric field will be induced which is non-conservative. And, as before, the wire has negligible resistance so the total electric field in the loop must be zero. But since \vec{E}=\vec{E_{c}}+\vec{E_{n}} , \vec{E_{c}} must be nonzero, and there must be an accumulation of charge on the terminals... oh wait, there are no terminals like before? So how is there an accumulation of charge, and where does the non-conservative electric field come from?! Now I'm confused!
Please can someone help me?
Cheers, Jon.
I have just started learning about inductors, and this is the method my book uses to show that Kirchhoff's second law is still valid even with inductors in a circuit, and to calculate the voltage drop across an inductor:
"According to Kirchhoff's loop rule, the algebraic sum of the potential differences around any closed circuit must be zero because the electric field produced by charges distributed around the circuit is conservative. We denote this such a conservative field as \vec{E_{c}}.
When an inductor is included in the circuit, the situation changes. The magnetically induced electric field within the coils of the inductor is not conservative. We denote this field as \vec{E_{n}}. We need to think very carefully about the roles of the various fields. Let's assume we are dealing with an inductor whose coils have negligible resistance. Then a hegligibly small electric field is requiresd to make charge move through the coils, so the total electric field \vec{E_{c}} + \vec{E_{n}} within the coils must be zero, even though neither field is individually zero. Because \vec{E_{c}} is nonzero, we know there have to be accumulations of charge on the terminals of the inductor and the surfaces of its conductors, to produce this field."
The author then goes on to use this, together with Faraday's law and the self-induced emf of the inductor, to work out the voltage drop across it. He concludes that we are fine in using Kirchhoff's second law, so long as we are defining the voltages in terms of the conservative part of the field.
I am trying to satisfy myself that this is true, but I have a problem: what if I apply the same thinking (regarding the total electric field being zero) to just a circular loop of wire in a changing magnetic field?
I know that an electric field will be induced which is non-conservative. And, as before, the wire has negligible resistance so the total electric field in the loop must be zero. But since \vec{E}=\vec{E_{c}}+\vec{E_{n}} , \vec{E_{c}} must be nonzero, and there must be an accumulation of charge on the terminals... oh wait, there are no terminals like before? So how is there an accumulation of charge, and where does the non-conservative electric field come from?! Now I'm confused!
Please can someone help me?
Cheers, Jon.
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