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Case 1

enter image description here

This is a very commonly discussed case in Electromagnetic Induction. In the case above, we need to find out the potential difference across the rod CD, in the presence of time-varying uniformly distributed cylindrical magnetic field as shown in the figure above.

Here we say that in equilibrium, the non-conservative electric field inside the rod completely balances the conservative electric field developed in rod, due to charge separation and the potential difference which we define in the ends of the rod, is due to the conservative electric field and not due to the non-conservative one, as there is no meaning of potential difference for non-conservative fields. I totally understand and agree with this explanation.

I do not have a doubt in the above case. The problem comes in the case below when we apply similar logic.

Case 2

enter image description here

Consider case 2 above. Here we have a similar time-varying, uniformly distributed cylindrical magnetic field in the region as shown. An equilateral triangular conductor is placed in the magnetic field, with its centroid coinciding with the centre of the cylindrical region. The three branches of the triangle have the SAME resistances. In this case, we wish to find out the potential difference between the points A and B. This is a very common question given in the text-books.

My question is on this case 2.

In this case, we are asked the potential difference between points A and B.

We know that potential difference is a concept associated with a conservative electric field and not with the non-conservative one. Now in this case, how the conservative electric field will come into existence? The free charges inside the conductor will simply start moving just by the effect of the non-conservative induced electric field and there is no need for us to introduce a conservative electric field in this case like an isolated rod in case 1. And, if there is no conservative electric field then there is no concept of the potential difference!

By this logic, either the question itself is wrong, that it is asking for the potential difference between points A and B, in spite of having no existence of the conservative electric field,

OR I am missing something.

Kindly help me.

How do we understand the idea of potential difference in Case 2?

EDIT1: In the triangular conductor, I have changed the resistances of all the sides to be the same.

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  • $\begingroup$ "OR I am missing something." - that the current through each branch is identical? $\endgroup$
    – Alfred Centauri
    Commented Jun 7, 2020 at 18:32
  • $\begingroup$ if a current flows in the circuit there is a conservative electric field.The non conservative field merely produces the emf for this $\endgroup$
    – Schwarz Kugelblitz
    Commented Jun 7, 2020 at 18:44
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    $\begingroup$ @Alfred : I understand the idea when the resistances are different then there will be accumulated charges on their interfaces and that would bring in a conservative electric field and that would change the potential differences across different resistances. MY doubt is different. Suppose all the resistances are same then there should not be any existence of conservative electric field and the idea of potential difference then loose the meaning, but it doesn't. How? $\endgroup$
    – Devansh Mittal
    Commented Jun 8, 2020 at 4:47
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    $\begingroup$ EDIT1: In the triangular conductor, I have changed the resistances of all the sides to be the same. $\endgroup$
    – Devansh Mittal
    Commented Jun 8, 2020 at 4:50
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    $\begingroup$ There would be conservative electric field set up within the loop provided there is current flowing through it.It is the potential difference of this field which they want you to calculate. $\endgroup$
    – Schwarz Kugelblitz
    Commented Jun 8, 2020 at 13:04

2 Answers 2

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And, if there is no conservative electric field then there is no concept of the potential difference!

But there is a conservative field.

To see this, imagine that the triangle is formed by three resistors (with the same resistance $R$ as per your EDIT 1) and assume the leads of the resistors are formed from ideal wire.

Something like this:

enter image description here

Image credit

There is, due to the changing magnetic flux threading the surface bounded by the resistors, a non-conservative electric field such that the line integral of this field (along the closed path defined by the resistors) is non-zero.

But the electric field within the ideal wire leads must be zero (even when there is a non-zero current through). So, it must be the case that their exists a charge distribution along the wire leads such that that the associated conservative electric field precisely cancels the induced electric field within the wire.

That is, just as in your case 1, the induced electric field separates charge against the resulting conservative electric field. If you think about this a bit, you'll see that this charge distribution along the leads is such that one end of the resistor body is more positive than the other end. Thus, there is a potential difference across the resistor (than can be measured if you're careful with the voltmeter lead placement).

Now can you generalize this to the case that the leads have non-zero resistance?

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  • $\begingroup$ What if the resistance is uniform throughout the whole triangular loop? $\endgroup$
    – user258881
    Commented Jun 8, 2020 at 19:09
  • $\begingroup$ @FakeMod, Note that I've asked OP if he can generalize to this. $\endgroup$
    – Alfred Centauri
    Commented Jun 8, 2020 at 19:18
  • $\begingroup$ According to my understanding, the conservative electric field should disappear. I don't want you to just directly give it to the OP, but I think either confirming or refuting my understanding should not be much of a spoiler for the OP. $\endgroup$
    – user258881
    Commented Jun 8, 2020 at 19:32
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    $\begingroup$ I've deleted a number of mildly inappropriate comments and/or responses to them. $\endgroup$
    – David Z
    Commented Jun 12, 2020 at 23:03
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    $\begingroup$ I finally understand to a good extent what you actually meant in the above response. Thanks a lot for it. $\endgroup$
    – Devansh Mittal
    Commented Jul 14, 2020 at 14:39
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Due to the time changing magnetic field there is induced non-conservative electric field and the line integral of that field along any closed path is not zero in accordance with Faraday's law. But what is interesting to note is that, if the assumption, that current in the circuit is not changing with time is made for such a case the problem reduces to classic ones. Sine current is not changing the drift velocity remains constant, the work done by the non-conservative electric field must appear in other forms (Conservation Of Energy(Not just mechanical)). It goes into heating the resistors. On doing calculations relating the work done in one cycle by the non-conservative electric field and heat dissipated by the resistors, one obtains the same equations as KVL but the I*R term now corresponds to energy dissipated by a unit charge in going through the resistor and not the potential drop. What we are doing is extrapolating the concept and call this term the potential drop but it actually isn't.

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