What is the ratio of the voltmeter readings in this E-field scenario?

  • Thread starter Septim
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In summary: This difference in voltage is called an emf. In the context of the problem, the emf is induced by the magnetic field on the outer loop (which includes the voltmeter). The emf causes a current to flow through the resistor, R2, from the upper to the lower terminal. The potential difference on the two terminals is the emf multiplied by the resistance of the resistor.
  • #1
Septim
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Greetings everyone,

While I was studying on my own using MIT OCW, I came across the following document. In that document in the last part you are asked to calculate the ratio of the reading of the two voltmeters positioned to the left and right of the loop. I did some work on them but I am unsure if I interpreted the question correctly. Any help would be appreciated.

Thanks,

P.S: I found it appropriate to post it in this section since it is related with classical physics and it is not a homework question.
 

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  • #2
Thanks for sharing this interesting problem. According to my calculation, the ratio becomes -R2/(2R1+R2).
 
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  • #3
Shouldn't it change when the loop goes ne times around the magnetic field can you write the Kirchhoff's Loop Rule equations for the loops?

Thanks
 
  • #4
Septim said:
Shouldn't it change when the loop goes ne times around the magnetic field can you write the Kirchhoff's Loop Rule equations for the loops?

Thanks

It's simple in fact. Did you take account of the emf induced on the outer loop?
 
  • #5
The outer loop that includes V_1 and the middle loop only, am I wrong ?
 
  • #6
Here are the equations

I=ε/(R1+R2)

V2=-R2I=-R2ε/(R1+R2)

V1=ε+R1I=(2R1+R2)ε/(R1+R2)

V2/V1=--R2/(2R1+R2)
 
  • #7
Suppose the magnetic field is coming out of the plane as it is decreasing. This induces ε volt emf which causes a current passing through R2 from A terminal to D terminal. The same emf is induced in the outer loop only that the current is negligible due to the high resistance of the voltmeter. The potential of the upper terminal will be ε volt higher than the potential of terminal D .

I hope its clear.
 
  • #8
Thank you I think it is clear now! In my expression V1/V2 = (ε + IR1)/IR2, and it gives the same answer if I plug in the approximated value of I into it. Then this is some sort of transformer if I am not mistaken. If the loop was wrapped N times the coefficient of R1 in the denominator of your expression would be (n+1), am I right ?
 
  • #9
Septim said:
Thank you I think it is clear now! In my expression V1/V2 = (ε + IR1)/IR2, and it gives the same answer if I plug in the approximated value of I into it. Then this is some sort of transformer if I am not mistaken. If the loop was wrapped N times the coefficient of R1 in the denominator of your expression would be (n+1), am I right ?

When wrapped N times, V1/V2=- (Nε + IR1)/IR2=-(N+1)R1/R2+N.

I am hesitant to call it a transformer because in a transformer we have input and output terminals. Where is the input terminal here? Also his ratio was calculated with the assumption than voltmeters have impedance much higher than R1 and R2. If we replace , for example V2, with a voltage source, things will change. Beside, although the ratio is independent of the flux, the voltages do depend on it.
 
  • #10
Thanks for the reply but I think the ratio would be ((n+1)R1+nR2)/R2.
 
  • #11
Septim said:
Thanks for the reply but I think the ratio would be ((n+1)R1+nR2)/R2.

It's the same. ((n+1)R1+nR2)/R2= (n+1)R1/R2+nR2/R2=(n+1)R1/R2+n
 
  • #12
Sorry my bad, I am indebted.
 
  • #13
Btw. the lecture that goes with that pdf is on youtube.
 
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  • #14
DrZoidberg said:
Btw. the lecture that goes with that pdf is on youtube.


Thanks a lot. That was entertaining as well as informative.

The main point of such a problem is that in a time-varying magnetic field, the voltage difference between two points in a conductive material ( here wires) can be non-zero.
 
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1. What is a non-conservative electric field?

A non-conservative electric field is a type of electric field that does not follow the principles of conservative forces. This means that the work done by the electric field on a charged particle depends on the path taken by the particle, rather than just its initial and final positions.

2. How is a non-conservative electric field different from a conservative electric field?

A conservative electric field is a type of electric field where the work done by the field on a charged particle only depends on the initial and final positions of the particle, and not on the path taken. In contrast, a non-conservative electric field does not follow this principle and the work done on a charged particle will vary depending on the path taken.

3. What are some examples of non-conservative electric fields?

One example of a non-conservative electric field is an electric field created by a changing magnetic field, also known as an induced electric field. Other examples include electric fields that are created by non-static sources, such as a moving charged particle or a time-varying current in a circuit.

4. How is the concept of potential energy related to non-conservative electric fields?

In a conservative electric field, the potential energy of a charged particle is conserved and only depends on its position in the field. However, in a non-conservative electric field, the potential energy of a charged particle will vary depending on the path taken. This is because the work done by the electric field on the particle is not constant.

5. What are some real-world applications of non-conservative electric fields?

Non-conservative electric fields have a variety of real-world applications, including in the generation of electricity through electromagnetic induction, in the operation of electric motors, and in the transmission of electrical signals through wires. They are also important in understanding the behavior of charged particles in plasma, which has applications in fields such as fusion energy research and space physics.

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