2

So,i was studying from some lecture notes from MIT's open course program,and i stumbled across this example enter image description here

The example says:The solenoid is so long that its external magnetic field is negligible. Its cross section is 20 cm^2 in area, and the field inside is to the right and increasing at the rate of 10^-2 T/s . Two identical voltmeters are connected as shown to points A and C on the loop, which encloses the solenoid and contains the two resistors of 50 Ohm each one. This gives us the readings of VM1=-10μV and VM2=10μV(calculations will be shown afterwards).Now,this is kind of weird,as the two voltmeters give us different measurements for the same point,but this is all included in the theory of induction due to changing magnetic flux(the integral of E.dS depends on the actual path).Things get weirder in this situation: I want to change the resistors to study what changes to the voltmeter's readings.I make R1=40Ω and R2=60Ω.
This gives us:enter image description here
For VM1 we consider the loop that encloses R1 and VM1 and it gives us a reading of IR1=0.2*40=8μV and the first voltmeter is showing -8μV because the voltage at A is bigger than that of C(as shown in the image)because of the direction of the flow of the current. But VM2 gives us IR2=-12μV(the negative sign due to the flow again).
Now,this i can not understand!We measure the same points,but we have different results!It is one thing to just get an opposite sign,but another thing to get two totally different numbers!
Can anybody please explain this to me?

Qmechanic
  • 201,751
TheQuantumMan
  • 7,664
  • There is a lecture by Prof Lewin that's online, and there was a question about that earlier. I will see if I can find the link - I believe this is essentially a duplicate of that one. – Floris Apr 02 '15 at 14:28
  • See this earlier answer and links therein. I haven't found the original one I'm looking for yet. But yes, it's Lewin... – Floris Apr 02 '15 at 14:32
  • I will see Lewin's lecture first.It might help – TheQuantumMan Apr 02 '15 at 14:36
  • It is to do with the flux enclosed in the loop. I asked a question on a similar topic earlier http://physics.stackexchange.com/questions/169076/em-induction-sliding-rod-with-resistance Look at my comment to Floris's answer. – Quantum spaghettification Apr 02 '15 at 15:28

1 Answers1

2

The key to this question is the position of the voltmeter and the flux it contains. Consider the diagram below: enter image description here

Let us say we are finding the voltage between A and D using the volt meter $M_1$. We can do this using one of two loops:

  1. One going straight from A to D. enter image description here
  2. The one going from A to B to C then to D. enter image description here

The first one will give an answer of $V_1=-IR_1$ since there is no flux passing thru the circuit we are considering (that including M1 and $R_1$). When we do it around the other loop then we must include the induced emf. So using Kirchoff's voltage law around this loop [$M_1$ and $R_2$] (which is valid if we include the induced emf). Then we get: $$-V_1+IR_2=\epsilon$$ but $\epsilon=I(R_1+R_2)$ so: $$-V_1=IR_1$$ we get the answer we got before. An analogous process can be done for the voltmeter on the second side. Such that: $$V_2=IR_2$$ [$V_1$ denotes the reading on voltmeter $M_1$ and $V_2$ that on $M_2$] The point is when we have non-conservitavie fields and when stating the voltage difference between two points we must also state where the voltmeter is positioned, the answer depends on its location, due to the induced emf. Note that for any loop we go round for a given voltmeter location then they all agree on the location (it would be very strange if we got a contradiction here). I hope this answers your question.

[P.s. Sorry for the poor diagrams, the circle in the middle is ment to be a solinode.]

Community
  • 1
Quantum spaghettification
  • 7,585
  • Your example is about how to measure the voltage difference in different ways.But it always give the same example.My problem is when two voltmeters measure different voltage difference for the same two points.I don't know if i dont understand something that you are trying to say.I am now watching the video that you told me in the comments – TheQuantumMan Apr 02 '15 at 15:50
  • The point is for a conservative field, voltage difference is poorly defined. The circuit that you are measuring the voltage with (i.e. that containing the voltmeter) is also affected by the induced emf in the circuit. My answer shows why you do get the values you do. The reason why with different voltmeter you get different answers is because each is differently affected by the flux passing thru the loop. – Quantum spaghettification Apr 02 '15 at 15:54
  • I understand what you are saying,but i can not see in your example where the two voltmeters measure the potential difference between the same two point but have different readings – TheQuantumMan Apr 02 '15 at 16:02
  • They have different readings since for $M_2$ the loop going thru $R_2$ contains no flux whilst that thru $R_1$ does. The flux thru the circuit changes the voltage measured across the two terminals. It is the other way round for $M_1$ the loop thru $R_1$ has no flux thru it, whilst that thru $R_2$ does. – Quantum spaghettification Apr 02 '15 at 16:05
  • So for the point A and D,M1 measures V=-IR1 while M2 measures V=IR2? – TheQuantumMan Apr 02 '15 at 16:10
  • Yes [the points $A$ and $B$ are equivalent, they could have been drawn at the same place. Lickwise with $D$ and $C$] – Quantum spaghettification Apr 02 '15 at 16:11
  • Ok,now i understand.But what is the intuition behind it?The voltmeters give us different potentials,so which one is true?I mean,we are talking about energy per charge!How can we get two different results?The mathematics prove it,but how can we think of it? – TheQuantumMan Apr 02 '15 at 16:19
  • For the larger loops (the one containing the flux) the electron is given an extra resultant 'shut', which increases or decreases its energy depending on the direction it is in. This does not occur in the path containing no flux. We thus need to include it this extra energy in the path containing flux but not in that not containing flux. – Quantum spaghettification Apr 02 '15 at 16:47
  • As i understood it from Lewin's lecture,it is like Kirchoff,but now if you integrate around the circuit(closed loop) you do not get zero(energy is not conserved) but you get the induced emf – TheQuantumMan Apr 02 '15 at 16:56