3
$\begingroup$

Consider the following two cases,

Case 1 – A coil moving through a constant magnetic field

Case 1 – A coil moving through a constant magnetic field

Case 2 – A conductor moving through a constant magnetic field enter image description here


Now, in Case 1 (a coil moving through a constant magnetic field) the induced current is zero, whilst in case 2, there is an induced EMF.

Is the reason that there is no induced current in case 1 because current gets induced equally on either side of the coil and it therefore cancels out. This would be concordant with case 2, as in case 2 a conductor likewise moves through a constant magnetic field and is able to induce an EMF.

OR

Is the reason that there is no induced current in case 1 because there is no change in flux. This appears to not be in line with case 2, as in case to, there is likewise 'no change in flux' but an EMF is induced.

Thanks

$\endgroup$
  • $\begingroup$ What is $F$ and $F_e$ in case two drawing? $\endgroup$ – Andre Maizel Jul 30 '15 at 0:34
  • $\begingroup$ Oh nothing relevant, I just pulled them of the net (essentially one is the force on a positive particle (F) and the other (Fe) is the force on an electron) $\endgroup$ – Pravin Jul 30 '15 at 0:39
  • $\begingroup$ In the second image... Shouldn't $F_e$ be in the upward direction? $\endgroup$ – Pranav Jul 30 '15 at 1:22
1
$\begingroup$

Treat the individual electrons in each conductor as if they were in a closed container, and there was otherwise a vacuum in that container. When moving through the magnetic field in the top picture (the ring), I would expect electrons to move to the bottom of the ring, leaving a net positive charge at the top. This will only occur until the electric force of repulsion of the electrons balances the magnetic force that is driving them to the bottom of the ring, meaning that there should be a potential difference between the top and bottom of the ring, but no current flow. The same analysis applies to the bar in the bottom photo, except the electrons would migrate to the top of the bar because it is going in the opposite direction. This means that there is an induced EMF in both cases.

Or, for a slight variation on this theme, treat the ring as a fat bar with a large hole drilled into it. There will be an induced EMF whether the hole is there or not, and there will also be no current flow in both cases.

$\endgroup$
  • $\begingroup$ That is an excellent answer David. Unfortunately I don't have enough reputation to up your answer. Thank you. $\endgroup$ – Pravin Jul 30 '15 at 0:46
  • $\begingroup$ I don't think there would be any EMF induced in both cases, or you would violate Faraday's Induction law. How can I satisfy rot(E) = -dB/dt in this case? $\endgroup$ – Andre Maizel Jul 30 '15 at 0:52
  • $\begingroup$ Andre, will another source convince you? $\endgroup$ – David White Jul 30 '15 at 1:10
  • $\begingroup$ It's incorrect to say that there is emf in the ring... Doesn't satisfy the - d(phy) /dt $\endgroup$ – Pranav Jul 30 '15 at 1:17
  • 1
    $\begingroup$ @DavidWhite: I think EMF is the wrong term to use in this situation. There will be a potential difference between the tops and bottoms of the ring and bar due to the Lorentz force, but there will be no sustained current flow because $dB/dt$ is zero. Electrical potential and EMF are slightly different: potential is between two points; EMF is the potential gained through a circuit, i.e., a closed-loop path. $\endgroup$ – Mark H Jul 30 '15 at 1:52
3
$\begingroup$

Clarification from another source:

Metal bar dragged through a magnetic field

Source: Physics For Scientists And Engineers, Paul A. Tipler and Gene Mosca, Sixth Edition, W. H. Freeman and Company, New York, 2008, p. 971, Fig. 28-20. I maintain that the loop will act the same as the bar. In other words, if you cut a thin slit down the center of the bar and less than the length of the bar (you leave it as one bar, not two,) the situations are exactly equivalent.

$\endgroup$
0
$\begingroup$

What I think is that there is no net emf induced in the first case and hence no current in the loop but there is emf induced in the second case but as the circuit is not complete, there is no current. Here is my reasoning: case 1 in the first case the flux which is (B. A) is not changing and hence emf I. E. d(flux) /dt is 0. One can visualise this in the following manner: consider the ring to be broken in two semicircles... One on the right hand side and one on the left hand side. In each consider the electrons present in each one... The electrons on the rhs ring move down due to the lorenz force acting on them and hence a negative charge appears at the bottom on this semicircle and a positive one appears on the other side... Lets replace the appearance of charges with a battery whose negative terminal is on the side where the negative charge developed and the positive side is where the positive charge developed. Follow the same procedure for the ring on the left hand side and replace the appearance of charges with a battery for convenience sake. Now join both the semicircles... You will find that the positive terminals of the battery are connected together and the negative ones are joined together... Hence the net emf generated in the ring will be zero. Hence no current flows in the ring. case 2 Consider your second case... The same logic here too... Due to the motion of the conductor in a magnetic field... The electrons under the force (q)(v x B) move to the upper end (the diagram is a bit incorrect there) of the rod developing a negative charge there, similarly, there is a development of positive charge in the other end which consequently leads to the development of an electric field which at some stage leads to a net 0 lorenz force . You can replace the development of charges with a battery for convenience sake. Hence you find that an emf is developed but as the circuit is incomplete, there is no current. If we were to connect the upper and the lower ends of the rod to an external circuit, that rod will help us induce a current in the external circuit as long as it's motion continues in the magnetic field. Note that if the upper and lower ends of the ring are connected to an external circuit, it also will act as a source of emf (in that case there would be two batteries in parallel (because of the two semicircular rods)).

$\endgroup$

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.