0
$\begingroup$

enter image description here

Here are two infinite wire with current I flowing through and I changes with time $I=kt+I_0$. They are $3a$ apart from each other and a square metal loop with length $a$ is placed at $a$ from one of the wire. The illustration is shown above.

It is easy to calculate the magnetic field from the Ampere's law along with the superposition principle.

For the magnetic field induced by a single wire, it is $B=\mu_0 I/2\pi r.$

The corresponding magnetic flux is $\phi=a \int_a^{2 a} \mu_0 I/2\pi r d r =a \mu_0 I \ln2/2\pi $ For two such wires, the total amount is $\phi=2 a \mu_0 I \ln2/2\pi $.

Then the electromotive force is $\varepsilon = -\frac{d \phi}{dt}= -a \mu_0 k \ln2/\pi$

The electrical resistance in AB, BC is $R_1$, and the electrical resistance in AD, DC is $R_2$. $R_1>R_2$.

According to the Ohm's law, the current in ABCD is $a \mu_0 k \ln2/\pi(R_1+R_2)$ The potential difference is $R_1 a \mu_0 k \ln2/\pi(R_1+R_2)$ from ABC, which is different from the potential difference of ADC,$R_2 a \mu_0 k \ln2/\pi(R_1+R_2)$.

So my question is what is the real potential difference between A and C, since a voltmeter must have a value if it is connected to this two points ?

Improve this question
$\endgroup$
5
  • $\begingroup$ "what is the real potential difference" - what do you mean by real? Before you answer, note that you can connect the leads of a voltmeter together (should give zero volts since the probes are connected to the same point, right?) and get a non-zero voltage that depends on the rate of change of magnetic flux threading the loop formed by the leads and voltmeter, i.e., the voltage read will vary simply by changing the orientation etc. of the leads. $\endgroup$ – Alfred Centauri Nov 17 '17 at 1:31
  • $\begingroup$ The real potential difference I written here is value you can read from a voltmeter and the rate of change of magnetic flux is k, a constant. This question comes from one of my exercises and it asks for the potential difference between A and B. However in my understanding, the potential difference is only related to the static electric field because its curl is always zero. From $J=\sigma f$, we can give the current and the path integral of $f$,V, then we have the ohm’s law,$IR=V$. $\endgroup$ – ukikaze Nov 17 '17 at 3:39
  • $\begingroup$ If V is value required, then it depends on the path you choose. If you connect a voltmeter to these two points, then it should depends on the way you connect the voltmeter to the loop. So I don’t know how to solve this regarding the question it asks. $\endgroup$ – ukikaze Nov 17 '17 at 3:39
  • $\begingroup$ Hint: The electrostatic potential is a quantity defined in the field of electrostatics, but your scenario is not a static one. $\endgroup$ – The Photon Nov 17 '17 at 5:22
  • $\begingroup$ @ThePhoton absolutely you are right the potential difference is only valid in electrostatic. Here the induced electrical field arises which makes electrostatic disappear. However if a voltmeter is connected to these two points there must be a value on it. So what’s the value, will it depends on how we position the voltmeter and the wire used for connection? That means if we put the voltmeter inside the square circuit or outside it, the value will be different $\endgroup$ – ukikaze Nov 17 '17 at 8:34
0
$\begingroup$

The question about a voltmeter reading as stated has no answer unless you include a statement about the position of the voltmeter and the leads that connect it to points $A$ and $C$.
By connecting a voltmeter into the circuit you have introduced extra loops which have emf induced in them.

enter image description here

It is not a matter of "real" potential difference it is to do with the reading on a voltmeter.
In the diagram above all three voltmeter readings will be different.

For more about this I would suggest that you watch Walter Lewin's lecture from about 33 minutes and the notes that he produced.
Also Ramamurti Shankar's lecture from about 38 minutes is worth watching?

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for your answer. I hold the same view of this question that the value should depend on the position of voltmeter and the potential required in this question is meaningless. I think I should ask my teacher about this. $\endgroup$ – ukikaze Nov 19 '17 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.