0
$\begingroup$

A metal rod of length $l$ is placed (as shown in figure) in a cylindrical time varying magnetic field. Find the potential difference across it. enter image description here

My answer:

As electric fields as forming circles around the centre with magnitude $$\frac12r\frac{dB}{dt}$$ whose magnitude is same along any chord at a fix distance from centre: $$\Delta V=E.l=\frac12r\frac32at^{\frac12}.l$$ $$\Delta V=\frac34arlt^{\frac12}$$

My question: Is $\Delta V =V_M-V_N$ or $\Delta V=V_N-V_M$ $\bf OR$ equivalently which side is at high potential?

Possible

  • $(i)$ As Field lines are circular from $M$ to $N$, $M$ must be high potential.

  • $(ii)$ As this field line apply a force on electrons- in the rod- towards left, then $M$ must have excess $e^-$ and $N$ must have deficiency; so $N$ would be high potential as $M$ is $(+)$ and $N$ is $(-)$.

$\endgroup$
  • $\begingroup$ Did you check this one? $\endgroup$ – user22180 Aug 3 '14 at 14:51
  • $\begingroup$ @user22180 yes it only tells to calculate field(that too not completely, which I know) but doesn't tell about higher potential... $\endgroup$ – RE60K Aug 5 '14 at 8:50
  • 1
    $\begingroup$ That one tells you can't determine electric fields uniquely. Maybe you are taking some particular choice. If you tell on which choice your results are based, then it may be easy to tell about the higher potential. $\endgroup$ – user22180 Aug 5 '14 at 11:07
  • $\begingroup$ I think it was not the magnitude but the component of Electric field along the chord that remains constant though. Not remember surely. $\endgroup$ – Mann May 7 '15 at 16:16
2
+50
$\begingroup$

There is an electric field due to the changing magnetic field, which has direction left -> right.

Due to this field, there is an immediate charge seperation caused in the metal rod (-ve near M and +ve near N) which creates another electric field of equal magnitude and opposite direction to perfectly cancel the external electric field.

Thus the net electric field when going from M -> N is 0. Hence the potential difference is also 0. This is a very logical result considering the fact that the two points are connected by a perfect conductor (shorted).

You can always find the potential drop due to one of the electric fields. That should (must) give you potential drops of equal magnitude and opposite sign. But with this you can't say that one of the sides is at a higher potential. They are both at the same potential.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I agree with udiboy on this one. If you insert a conducting wire, you also break the translation symmetry along the z-axis, which means that the potential difference which you would measure at different sections of the magnet would depend on z. Far away from the wire it would be the "free problem", near the wire ends, the potential would be shorted. To make the problem symmetrical in z, one would have to look at an infinite conducting strip parallel to z. Such a strip would, of course, separate the two halves of the field due to eddy currents and destroy the homogeneity of the field. $\endgroup$ – CuriousOne Aug 13 '14 at 21:50
  • $\begingroup$ what if you join two ends by a conducting wire, wouldn't current flow. $\endgroup$ – RE60K Aug 14 '14 at 4:29
  • 1
    $\begingroup$ the another created field exists only if rod is not closed in a loop, if it is, there would no field be created and the electrons would move smoothly $\endgroup$ – RE60K Aug 14 '14 at 4:31

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.