Answer to the Question 02/97
The question was:
Flux of the magnetic field B through a metalic ring is changing
with time. It creates 12 Volt electromotive force in the ring.
A voltmeter is connected to the contacts shown in the picture, i.e.
the contacts are are separated by 1/4 of a circle.
What will be the reading of the voltmeter?
The problem was solved correctly by Itamar Borukhov, Eilon Brenner,
Boaz Kol
and Udi Fuchs from Tel Aviv U. (4/6/97).
The answer: if the voltmeter is positioned
in the lower-right corner outside the ring, then it will show 3V, while
if it is positioned in the upper-left corner outside the ring, then
it will show 9V.
The outline of the solution is as follows:
1. Since this is NOT an electrostatic problem it is impossible to define
potential differences (voltages). Neverthless the integral of the electric
field E along a certain segment (electromotive force - EMF)
of a ring plays a very similar role:
Since the Ohm's law is still valid (current is proportional to the local
field, therefore the sum of resistances of segments of circuit multiplied
by currents must be equal to the EMF in that circuit. However, the EMF
is equal to the rate of change of flux within that specific loop.
2. Voltmeter has high internal resistance an measures the current flowing
through it. It displays on its scale the current multiplied by internal
resistance (that's what we call "reading of the voltmeter").
3. The current is conserved at every junction.
4. One needs to choose two loops in the electric scheme and apply the above
laws.
Here is what Borukhov, Brenner, Kol and Fuchs actually wrote:
The Voltmeter in the picture will read 3V.
The answer depends on the geometry in which the Voltmeter is connected
to the circuit. For example, if the Voltmeter was connected to the same
points, but this time from the other side of the ring - it would read 9V.
The subtle point is that the voltage is not a local quantity but
rather a global one depending on the amount of flux circled by the
current loop. Therefore one has to solve for the current passing through
the Voltmeter.
One way to solve the problem is to look at closed current loops
using Kirhoff's law (see Figure). [Note: the "power sources" in the figure
are "symbolic" - they denote integrals of electric field.]
For example, suppose we want to solve for the currents using
the top and bottom loops. Then the equations would be:
top: 12V = 3/4*R*I1 + 1/4*R*(I1-Ir)
bottom: 0 = Ir*r + 1/4*R*(I2-I1)
which gives in the limit r>>R: Ir*r = 3V
If we want to use the top circuit and the external one then we get:
top: 12V = 3/4*R*I1 + 1/4*R*(I1-Ir)
external: 12V = 3/4*R*I1 + Ir*r
which gives the same result.
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