1
$\begingroup$

I came across this question in an exam. The answer to this was given as YES. But I think otherwise.

They say between A and B, there is zero potential difference.

But if i simply alter the diagram a bit (as you can see in the second diagram), then a net current would surely flow from A to B as the line AB doesn't divide the circle symmetrically.

And if there is a current it automatically means there IS a potential difference between A and B ! enter image description here

So, is my logic right ?(Mostly its wrong, so how ?)

$\endgroup$
2
$\begingroup$

Perhaps the easiest way to see that there can't be a potential difference between $A$ & $B$ is a symmetry argument. You're tempted to say that $A$ is at a higher potential than $B$ so that current will flow from $A$ to $B$. But continuing along the loop, I find that current must also flow from $B$ to $A$, which would lead me to conclude that $B$ is at a higher potential than $A$. There is nothing along the wire to cause a jump in the potential (e.g. a battery). $A$ cannot be at both a higher and lower potential than $B$, so they must be at the same potential: there is no potential difference.

$\endgroup$
2
+50
$\begingroup$

The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative.

The two pieces of the electric field $\mathbf{E}$ can be seen explicitly if you express the electric field in terms of the electric potential $\phi$ and the magnetic vector potential $\mathbf{A}$: $$\mathbf{E} = - \mathbf{\nabla} \phi - \partial_t \mathbf{A}.$$

The two pieces of the electric field are the conservative piece $- \mathbf{\nabla} \phi$, and the emf piece $ - \partial_t \mathbf{A}$. The first piece comes from charges and you are familiar with it from electrostatics. In your problem, there are no net charges, so $\phi$ is zero, the potential is the same everywhere, and there is no electric field originating from the existance of net charges. The other piece, $- \partial_t \mathbf{A}$ does not occur in electrostatics, and in fact only occurs when there are changing currents. It is probably new and unfamiliar to you, which is probably why you are confused. The electric field from your problem is entirely due to this piece (since we saw the other piece was zero).

So as you can see, even though you have a non-zero electric field, the electric potential $\phi$ is in fact zero (since there are no net charges), and so every point is at the same potential. The non-zero electric field is instead entirely due to a second piece given by $- \partial_t \mathbf{A}$.

$\endgroup$
1
$\begingroup$

Here, there is a time varying magnetic field at work, and it's flux through the given loop changes. Thus there is a non-Electrostatic Field induced along the wire, proportional to the rate of change of the flux. However, it is a non-electrostatic field", for example the closed integral over a path for this field isn't zero. Also, the line integral of this field, isn't called the "potential". Only the electrostatic Eelctric field has a potential. Hence, though there is a current flowing, there is no potential difference across any point.

On the other hand, the line integral of this field from A to B is non zero, and from B to A is also non zero and they don't add to zero either.

$\endgroup$
1
$\begingroup$

This is a very simple problem to tackle. The points A and B are connected by a conducting wire and nothing else. Therefore, there is no potential difference from A to B. The magnetic field is a red herring. We are not told otherwise, so we assume the wire is an ideal conductor. That means there is never a potential difference across it.

Remember Ohm's Law is $V=IR$. If $R=0$, then no matter what the current is, $V=0$.

$\endgroup$
1
$\begingroup$

In case of varying magnetic fields the electric field is considered as nonconservative. Thus, it is simply not possible to define a potential for induced electric field (See also this post with answers).

Alternatively, one can think of the induced current as charges moving in the circle of radius $r$. Through every part $r \,d\phi$ flows the same amount of charges every second since the change of the magnetic field is constant. Due to this flow there will be no accumulation of charges in any interval $r\, d\phi$ of the circle. The points $A$ and $B$ will feel the same amount of charges and thus no potential will be measured.

A slightly different problem with an induced current but not constant magnetic field change including the discussion of nonconservative fields has been lectured by Prof. Walter Lewin (min. 35). He shows that definition of potentials in an induced electric field is pointless.

$\endgroup$
0
$\begingroup$

Wait, is this just a constant magnetic field and conducting loop or is the loop merely an illustration?

Regardless, just use Stokes' theorem and Faraday's law to get $\oint \ \mathbf{E} \cdot d\mathbf{l} = 0$ for $\partial \mathbf{B}/\partial t = 0$.

The point is that electrostatic electric fields are conservative (though electrostatic waves can be slightly different, but that is an unnecessary complication here), or in other words, the path does not matter.

$\endgroup$

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.