A little brain teaser

Ivan Seeking
Staff Emeritus
Science Advisor
Gold Member
7,093
174

Main Question or Discussion Point

Here is a favorite of mine from college that I thought you might enjoy: Consider a closed square loop of wire with two resistors, R1 and R2, each located in-line and symmetric about the center of the loop. Two voltmeters each measure the voltage change, V1 and V2, across R1 and R2 respectively. The loop is positioned such that a constantly increasing magnetic field vector [dB/dt > 0] passes through the center of the loop causing current i to flow.
````````i -->
|----------------|
|``````````````|
|``````````````|
R1```(dB/dT)```R2
|```````X``````|
|``````````````|
|-----------------

We demand that: The conductor, resistors, and voltmeters are perfect, and that R1 is not equal to R2.

Here's the problem, if R1 not = to R2, then according to Ohm's law, since i is constant over the loop, V1 not = V2. But the voltmeters are connected in parallel, so it seems they must read the same. So do they read differently or the same and why? Sorry, the diagram didn't quite work.

I know this won't be hard for everyone.
 
Last edited:

Answers and Replies

marcus
Science Advisor
Gold Member
Dearly Missed
24,712
783
Nice problem

In conventional circuit analysis one habitually assumes
that voltage is constant along a perfect conductor
(same at either end of each segment of wire)...

I see the answer but will not say, so other people
can enjoy the brain-teasing.
 
916
0
The question is that is not a parallel circuit but a series circuit. In a parallel circuit, the voltage must be applied at opposite sides of the circuit, and then the two resistors can share the voltage. In the case of your example, the varying magnetic field is causing an electromotive force, but this electromotive force appears at each point of the circuit, so, its like the source of voltage its in every point of the circuit, making thus a series circuit
 
508
0
Originally posted by Ivan Seeking
But the voltmeters are connected in parallel, so it seems they must read the same.
Even if R1=R2, the readings will not be the same, but will be +V and -V.
 
chroot
Staff Emeritus
Science Advisor
Gold Member
10,165
34
The voltage is, of course, not constant along the wires. The voltmeters will indeed measure different voltages.

- Warren
 
marcus
Science Advisor
Gold Member
Dearly Missed
24,712
783
Originally posted by chroot
The voltage is, of course, not constant along the wires. The voltmeters will indeed measure different voltages.

- Warren
Correct. Maybe we should have a physic game of Q/A analogous to the "astronomy" thread. In which case it would now be chroot's turn to pose a not-to-hard question to which he knows the answer.


A favorite version of the question chroot just answered is the one about the airplane flying the polar route. It is going 200 meters per second and its wingspan is 30 meters----what is the voltage difference between the two wingtips? I am not asking this question because it is a cousin of the one asked by Ivan, just recalling it. the plane is in a region where the earth's magn. field is roughly vertical and of such and such a strength etc etc.

Chroot do you have an intriguing but easy physics problem?
 
chroot
Staff Emeritus
Science Advisor
Gold Member
10,165
34
Actually, I rather like the idea of a Physics Q/A game... I'll start another thread for it. Your quetion is a good starting point!

- Warren
 
marcus
Science Advisor
Gold Member
Dearly Missed
24,712
783
Originally posted by chroot
Actually, I rather like the idea of a Physics Q/A game... I'll start another thread for it. Your question is a good starting point!

- Warren
that is fine. use the problem without attribution---I dimly remember it
from some physics textbook----I dont recall the strength of the geomagnetic field where it is vertical
something like 1/2 gauss? how many microTesla?----should have it in Tesla because everything is SI these days.
If you decide to use the problem you had best formulate it with your own words and numbers.

half a gauss is 50 microtesla?
 
Ivan Seeking
Staff Emeritus
Science Advisor
Gold Member
7,093
174
A little note on this one: I once posed this paradox to a burly old Ham guy - a salt of the earth, old school low tech type with thumbs twice the diameter of mine. He crossed his arms, raised an eyebrow and then declared with great authority: "I believe in Ohm's Law!"
 
marcus
Science Advisor
Gold Member
Dearly Missed
24,712
783
Originally posted by Ivan Seeking
A little note on this one: I once posed this paradox to a burly old Ham guy - a salt of the earth, old school low tech type with thumbs twice the diameter of mine. He crossed his arms, raised an eyebrow and then declared with great authority: "I believe in Ohm's Law!"
credo in unum deum
patrem omnipotentem
factorem coeli et terrae
visibilium omnium
et invisibilium
except when there is a changing magnetic field
 
398
0
Here's one part of it.

We can't just consider the magnetic field. The changing magnetic field induces an electric field and we have to include that. The resistors are dissipative and power has to go into them, so that means there has to be a Poynting Vector involved.
 
Alexander
Originally posted by Ivan Seeking


Here's the problem, if R1 not = to R2, then according to Ohm's law, since i is constant over the loop, V1 not = V2. But the voltmeters are connected in parallel, so it seems they must read the same. So do they read differently or the same and why?

I know this won't be hard for everyone.
Assuming that resistors and voltmeters are small in length (compared to the length of the rest of the circuit) then V1 =-R1V/(R1+R2),and V2 =+R2V/(R1+R2), where V=-A(dB/dt), A = loop area. Note that a wire is not equipotential in a varying with time B field.
 
Last edited by a moderator:
Ivan Seeking
Staff Emeritus
Science Advisor
Gold Member
7,093
174
Originally posted by Alexander
Assuming that resistors and voltmeters are small in length (compared to the length of the rest of the circuit)
True! But don't worry, my perfect voltmeter, leads, and resistors, are not subject to the laws of induction.


Note that a wire is not equipotential in a varying with time B field.
I was wondering if you were adding something or just restating it for clarity. Chroot already gave this answer:
"The voltage is, of course, not constant along the wires. The voltmeters will indeed measure different voltages.

- Warren "
 
398
0
Here's the answer

found by using a little substitution and separation.

The voltage dropped across both resistors has to be equal and opposite to the voltage developed across all four sides of the square. A condition was that the current through the loop was I.

Therefore Vtotal = I(r1 + r2)

and each side of the square developes 1/4 of the total voltage.

So taking our measurements from the corners the meters read,

I(r1−{r1+r2}/4)

and

I(r2−{r1+r2}/4)
 
Last edited:
Alexander
You forgot that a voltmeter has leads (=sources of induced voltage).
 
398
0
But only if

Originally posted by Alexander
You forgot that a voltmeter has leads (=sources of induced voltage).
they encircle the changing magnetic field. Don't forget the topological nature of a magnetic field.
 
Alexander
How can you "encircle" two distant points? And what is "topological nature" of magnetic field?
 
398
0
When you hook a meter up

Originally posted by Alexander
How can you "encircle" two distant points? And what is "topological nature" of magnetic field?
to an electrical circuit you complete a closed path, you have two circuits. Draw out the picture and you'll see what I mean.

And the second item, magnetic field lines don't close, they always circle back. There are some pretty good write ups about the topology of the magnetic field, the Ryder book on Quantum Field Theory has one.
 

Related Threads for: A little brain teaser

  • Last Post
Replies
2
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
Top