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

[Septim]

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.

 

[Hassan2]

Thanks for sharing this interesting problem. According to my calculation, the ratio becomes -R2/(2R1+R2).

 

[Septim]

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

 

[Hassan2]

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?

 

[Septim]

The outer loop that includes V_1 and the middle loop only, am I wrong ?

 

[Hassan2]

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)

 

[Hassan2]

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.

 

[Septim]

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 ?

 

[Hassan2]

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.

 

[Septim]

Thanks for the reply but I think the ratio would be ((n+1)R1+nR2)/R2.

 

[Hassan2]

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

 

[Septim]

Sorry my bad, I am indebted.

 

[DrZoidberg]

Btw. the lecture that goes with that pdf is on youtube.

 

[Hassan2]

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.