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Consider this situation in which a uniform cross section conducting neutral ring is placed in a changing magnetic field which is given by $B=kt(-\hat{k})$ ($k$ is a constant and $t$ is time).

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The upper half has a resistivity twice that of lower half. Now, according to Faraday's law of electromagnetic induction, the net emf enduced in the ring will be of magnitude $k\pi r^2.$ Due to this emf, a net current will flow through the uniform cross section of the ring which will be equal for lower as well as upper half hence current density $J$ will be equal in both halves. So, (Electric field)/(resistivity)=constant (from the equation $J=\sigma E$), but since the resistivity in lower and upper halfs are different, the electric field must be different, too. But, in such cases, the non-conservative electric field is given by the formula $= -(r/2)\times \mathrm{d}\Phi_b/\mathrm{d}t$ (rate of change of magnetic flux), which is same for both halves. How does the electric field change in both halves? Is there a conservative electric field induced?

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  • $\begingroup$ Break up the very long sentence beginning with “Now” $\endgroup$
    – Bob D
    Commented May 17 at 16:28
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    $\begingroup$ please suggest the edit, i will approve it if need $\endgroup$
    – FUSION X
    Commented May 17 at 16:30
  • $\begingroup$ This is a question about concepts. It's not asking for help with a homework-type task. $\endgroup$
    – Philip Wood
    Commented May 17 at 21:07

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"Is there a conservative electric field induced?" Yes, I think so.

The magnetic field, directed into the page and steadily increasing in time, generates a non-conservative electric field in an anticlockwise direction around the loop, and a conventional current in the same sense. Electrons, travelling clockwise, will tend to pile up at A as they encounter more resistance in the top half. There will be some depletion of electrons at B.

The conservative field due to these effectively static charges will boost the non-conservative field in the top half, where the resistivity is higher, but diminish the resultant field in the lower half, where the resistivity is lower. So once a steady state is reached, the current density in the two halves will be the same, just as your argument shows that it should be.

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  • $\begingroup$ i did not understand how the electrons get piled up at the junction, if you can provide any reference from any source about such behaviour of electron in any electrical circuit then it would be greatly appreciated. $\endgroup$
    – FUSION X
    Commented May 17 at 18:11
  • $\begingroup$ @FUSIONX this can be seen from Ohm's law. If current density $\mathbf j$ is stationary (steady state of current), and there is a rapid jump in conductivity when crossing junctions A,B, there has to be a jump in electric field at those junctions. Mathematically, Ohm's law says $\mathbf j =\sigma \mathbf E$, so a jump in $\sigma$ has to be cancelled by opposite jump in $\mathbf E$, in order for $\mathbf j$ to not have a jump there. A jump in $\mathbf E$ has to be associated with a non-zero charge accumulation, due to the Gauss law for electric field: $\nabla\cdot\mathbf E = \rho/\epsilon_0$. $\endgroup$
    – Ján Lalinský
    Commented May 17 at 19:49
  • $\begingroup$ @FUSION X I'm going to have to let you down; the 'piling up' was based on intuition. Yet, as you have seen, and as supported by Ján Lalinský's comment, it has to be so. $\endgroup$
    – Philip Wood
    Commented May 18 at 10:46
  • $\begingroup$ I got it.thankyou for the explanation @Ján Lalinský $\endgroup$
    – FUSION X
    Commented May 19 at 17:52

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