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I was reading the Griffin's Electrodynamics textbook and I came up with this question:

enter image description here

Suppose we have a uniform time-dependent magnetic field, I know that by Faraday's Law, we can calculate the induced electric field at the dashed Amperian loop to be $E=-\frac{s}{2}\frac{dB}{dt}\hat{\phi}$. However, if we choose another Amperian loop that intersects with the previous loop, then at the two intersections (the two blue dots), Faraday's Law will give a different direction of the induced electric field. May I ask how should I understand this difference?

My current thought is the induced electric field won't make physical sense if we don't have an actual "receiptor". That is to say the actual direction and magnitude of the electric field would be dependent on the shape of the actual circuit I put into the magnetic field. Therefore, the two imaginary loops in my question actually correspond to two different physical questions. Therefore the induced electric field will be different. Is my thinking correct?

Qmechanic
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EricChester
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  • You should illustrate this new loop, because it isn't clear what you mean, the induced E field however is not dependent on loop shape but it would be easier to illustrate if you take us through your example with an illustration. – Triatticus Jan 23 '23 at 19:55

  • "we can calculate the induced electric field at the dashed Amperian loop to be ..." We can't. There is no unique way to infer electric field from the magnetic field in this scenario. See my answer here https://physics.stackexchange.com/a/746838/31895 .

     – Ján Lalinský Jan 23 '23 at 20:51
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    In a realistic scenario, magnetic field is not uniform everywhere but decays with distance, and electric field is unique, even though we can't determine it just from magnetic field alone. Distribution of sources (charges and currents) in space matters. – Ján Lalinský Jan 23 '23 at 20:53
  • @Triatticus Hi, I've just added the illustration. Thank you! – EricChester Jan 24 '23 at 03:53
  • @JánLalinský Thank you so much! I think that post you mentioned is very similar to mine. I'll think more on that and if I still have more questions, I'll post it here. Thank you again! – EricChester Jan 24 '23 at 03:57

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Sorry this is not the answer but comment field is very limited then I write here.

If there is no wire circle in the uniformly varying magnetic B-fields, there are B-fields and electric E-fields (emf) in space. Although the B-fields are uniform over the space positions, the emf E-fields become some non-uniform distributions.

If a wire circle is placed at the "B-field center" so as to maintain axis symmetry, a uniform electric current $\vec{J}$ is induced inside the wire circle in accordance with the uniform E-field distribution along the wire.

If the wire circle is placed at some deviated point of the "B-field center", then the axis symmetry is broken and the tangent component of the emf E-field along the wire circle differs from place to place. In this case, $-\text{grad}\phi$ distribution is generated both inside wire and space, and electric current distribution becomes uniform along wire. Note that electric scalar potential $\phi$ is single valued function; no contributions to emf.

HEMMI
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  • Thank you for providing an explanation. Here is what I understand so far: In reality, there is no uniform magnetic field spanning the whole space. Therefore, we have to rely on the symmetry of the configuration if we want to use Faraday's law and assume a uniform electric field on the wanted circuit (the dashed circle for example). If there is no symmetry, then the E field on the circuit is no longer uniform. But inside the wire, E field has to be uniform to sustain a uniform current. Thus, -grad(V) is generated to smooth the current. And that is complicated to calculate. Is this correct? – EricChester Jan 25 '23 at 17:58
  • I completely agree with your written statement. I am sorry that it would have been difficult for you to understand due to my poor English. An interesting case is when a circularly connected wire is cut with nippers to make an open circuit. The $-\text{grad}V$ cancels the electric fields along the wire and the rotating current becomes zero. – HEMMI Jan 25 '23 at 21:46