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For example, in the case where there is a bar magnet moving through a loop of wire with some resistance, I know that by Lenz's law, the induced emf $\mathcal E$ in the circuit is equal to $-\dfrac{d\Phi_B}{dt}$.

But $$\int_C \mathbf{E}\:\cdot d\mathbf{l}$$ is equal to $-\Delta V$ by definition (where $\Delta V$ is the potential difference between the start and end locations.

Wouldn't this imply that $$\oint \mathbf{E}\: d\mathbf{l}= -\mathcal E = \frac{d\Phi_B}{dt}$$

since $\mathcal E$ is an induced potential difference?

According to Faraday's Law in integral form,

$\oint \mathbf{E}\: d\mathbf{l}$ is equal to $-\frac{d\Phi_B}{dt}$, and not positive $\frac{d\Phi_B}{dt}$

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them, as well as some mildly inappropriate comments. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering; and also remember our expectations for conduct on this site. $\endgroup$
    – David Z
    Commented May 8, 2020 at 22:26
  • $\begingroup$ Got your message John, no problem. :-) I wanted to suggest adding just a little information into the question to describe the physical situation you're asking about. For instance, does your question apply to any arbitrary circuit which is a closed loop? If so, it would be useful to say that you're talking about a closed circuit. Also, what kind of magnetic field is present? (spatially uniform, bar magnet, time-varying, etc.?) What is $\Delta V$? It might also help to be more explicit about what's wrong with the last equation, or why you find it a surprising result. $\endgroup$
    – David Z
    Commented May 8, 2020 at 22:41
  • $\begingroup$ ok I believe the questoin should be more clear now. $\endgroup$ Commented May 8, 2020 at 23:04
  • $\begingroup$ That's definitely better, yes. I think it would still be useful to say what $\Delta V$ is; as far as I can tell, you only use that symbol once and you don't explain what it means anywhere in the question. (I mean, it's fairly clear that it should be some kind of difference in electric potential, but a difference between what and what?) $\endgroup$
    – David Z
    Commented May 9, 2020 at 0:03
  • $\begingroup$ ok I added context for that $\endgroup$ Commented May 9, 2020 at 0:05

1 Answer 1

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The equation you have trouble with is meant for any closed path rigged with a prescribed sense of circulation in space:

$$ \oint_\Gamma \mathbf E\cdot d\mathbf l = -\frac{d\Phi}{dt} $$

The path $\Gamma$ has to be closed and the sense of circulation has to be prescribed for $\Phi$ to be defined.

This loop integral is actually determined by the induced electric field only, as the integral of the Coulomb field is zero. Therefore it is also called induced EMF for a given loop and sometimes denoted $\mathscr{E}$: $$ \mathscr{E} = \oint_\Gamma \mathbf E_i\cdot d\mathbf l $$ where $\mathbf E_i$ is induced electric field.

Induced EMF at time $t$ for an oriented closed curve $\Gamma$ is a work of induced electric field as it exists at time $t$ in the hypothetical process where unit positive charge moves along the path and completes one full circuit, while the electric field remains unchanged.

This EMF is (as a concept) not a potential difference. Potential difference, to be meaningful in practice, requires two points to be given, but the above equation does not require or imply such points.

If we additionally introduce two such points, then potential difference between them is meaningful concept, but in general it is not determined by induced EMF. Potential difference characterizes effect of the Coulomb electric field; induced EMF characterizes effect of induced electric field. In general, total electric field is neither purely Coulomb nor purely induced, but for low frequency phenomena, it can be understood as sum of the Coulomb part and the induced part.

Only in case the two points are connected by a highly conductive body, such as wire, and net electric field in the wire is negligible, can the potential difference be connected to the induced EMF. Zero electric field inside the wire means the Coulomb field at any point of the wire $\mathbf E_C$ is almost exactly opposite to the induced field $\mathbf E_i$. So potential difference magnitude is almost the same as magnitude of the induced EMF. However, they have opposite sense!

This is the case of ideal inductor. Let us isolate one inductor from any circuit. It has two endpoints:

                   __    __    __
                  /  \  /  \  /  \
            1 ---/    \/    \/    \--- 2

-> prescribed positive direction of this part of circuit ->

For these endpoints of ideal inductor, potential difference (also called voltage) is defined, as integral of the Coulomb field: $$ \text{Voltage drop from 1 towards 2} = V_1 - V_2 = \int_1^2 \mathbf E_C\cdot d\mathbf l $$

In the wire, Coulomb field is almost equal to minus the induced field. We can express the integral above as

$$ \int_1^2 -\mathbf E_i\cdot d\mathbf l. $$ This is, by definition, minus the induced EMF for path from 1, going through inside of the conductor to 2. The induced EMF is $$ \mathscr{E} = \int_1^2 \mathbf E_i\cdot d\mathbf l = -L\frac{dI}{dt}. $$ (In the last step, the induced EMF for the inductor is expressed using inductance of the inductor $L$ and electric current $I$).

Thus $$ \text{Voltage drop from 1 towards 2} = - \mathscr{E} = L\frac{dI}{dt}. $$

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