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Faraday's Law of induction states that the work done per unit charge by the (induced) electric force along a loop of wire, or the emf, is minus the rate of change of magnetic flux through the loop/surface and is given by the equation $$\mathcal{E}_{Ind.}=\oint \vec E_{Ind.}d\vec l=-\frac{d\Phi_B}{dt} \tag{1}$$

The minus sign is necessary because the induced current must travel in a way such that the induced magnetic field produces a change in flux that is opposite to that of the original magnetic field, so that energy is conserved. (Lenz's Law)

So according to Eq. $(1)$ when the magnetic flux through an arbitray loop increases, that is $$\frac{d\Phi_B}{dt}>0 \tag{2}$$ the induced emf $\mathcal{E}_{Ind.}<0$.

*But if $\mathcal{E}_{Ind.}$ is defined as "the work done per unit charge by the (induced) electric force along a loop of wire" or $W_{Ind.E}/q$, how can it ever be negative (assuming q to be positive) if the induced electric force is always acting along the direction of motion of the positive charges in the induced current?

*Shouldn't the work done by an electric force $W_E$ only be negative if the electric force is acting against the direction of motion (i.e. an external agent is pushing the positive charge from a region of lower potential to higher potential)?

I thought that in general, the $\mathcal{E}$ and V symbols for a circuit denote $\Delta V_{-\to +}$ (and hence = $W_{E,+ \to-}$ since $-W_{E, - \to +}=W_{E,+ \to-}$) since they always take positive values (apart from the reading of a voltmeter that's connected the wrong way around).

*So does this imply that the $\mathcal{E}_{Ind.}$ symbol in Faraday's Law is different to the $\mathcal{E}$ that appears in the cell of a circuit (because $\mathcal{E}_{Ind.}$ can be negative)?

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    $\begingroup$ \mathcal{E} ($\mathcal{E}$) would be better than \epsilon ($\epsilon$) for denoting EMF. $\endgroup$
    – G. Smith
    Commented Jul 21, 2019 at 6:08
  • $\begingroup$ Make your edits clear, as you can make answers look poor just because you changed your question - it is not the fault of any who answer if the question was not clear... $\endgroup$
    – user207455
    Commented Jul 21, 2019 at 7:24
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    $\begingroup$ @Solar Mike I thought the present title is not very different to the previous one and the content in the question has almost remained unaltered by any of my edits- the answer would only look poor if one just read the title and skipped to the answer straight away (hence why I edited the question title...but I do appreciate that you pointed it out). $\endgroup$
    – Chern-Simons
    Commented Jul 21, 2019 at 7:45

4 Answers 4

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This is where it's a good time to "converse with the math". Let's look at the equation:

$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$

which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:

$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$

Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.

Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.

But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.

(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)

Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).

Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.

From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.

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    $\begingroup$ "when E points opposite to dl." and isn't when the integral limits $l_2>l_1$ where $\int_{l_1}^{l_2}$ a necessary condition for $W_{electric.force}$ to be negative as well? $\endgroup$
    – Chern-Simons
    Commented Jul 21, 2019 at 9:45
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    $\begingroup$ @Contravariance : Well yes, if you flip the bounds, then you're in effect reversing the course of the imaginary charge through the loop, and hence when before the force would be working with you it will be working against, and conversely, so the sign of the work is also reversed. $\endgroup$
    – The_Sympathizer
    Commented Jul 21, 2019 at 9:51
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    $\begingroup$ So wouldn't Eq. (1) be more correct if it was written as $\mathcal{E}_{Ind.}=|-\frac{d \Phi_B}{dt}|$ (i.e. not just $|\mathcal{E}|=|-\frac{d \Phi_B}{dt}|)$ since Eq.(1) it's always referring to the the work done by $F_{electric.field}$ during the "actual movement of charges" which is positive since there's no "hand" grabbing the charge and fighting against the electric force in an actual electrical circuit? $\endgroup$
    – Chern-Simons
    Commented Jul 21, 2019 at 10:05
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    $\begingroup$ @Contravariance : The movement of an imaginary charge is what intuitively/conceptually defines the emf, which then becomes mathematically formalized via the line integral; Faraday's law describes how that quantity behaves in a physical system. If we interpret backwards through the intuitive idea, when $\mathcal{E}_\mathrm{ind}$ becomes negative, that means that at that point, were we to do that imaginary motion, negative work would be done. $\endgroup$
    – The_Sympathizer
    Commented Jul 21, 2019 at 10:08
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    $\begingroup$ @Contravariance : The physical relevance, by the way, is that when the emf reverses sign, the current is compelled to reverse direction, which is the physical reason we can build alternating-current power systems. If we defined it in terms of the direction of actual charge motion and not of this imaginary charge motion, we would be unable to make that distinction and EM theory would be that much more useless for doing that. The imaginary charge motion sets a reference direction. $\endgroup$
    – The_Sympathizer
    Commented Jul 21, 2019 at 10:39
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The minus sign is necessary because the induced current must travel in a way such that the induced magnetic field produces a change in flux that is opposite to that of the original magnetic field

The phrase in bold is incorrect as it could be in the same direction as that of the original magnetic field if the magnitude of the original magnetic field is decreasing.

Consider a simple series circuit consisting of a cell of emf $\mathcal E_{\rm cell} $, a switch and an inductor $L$.

At a time $t=0$, when the current is the circuit $i=0$, the switch is closed.

$\mathcal {E_{\rm cell}} + \mathcal E_{\rm induced}=0\Rightarrow \mathcal {E_{\rm cell}} - L \frac {di}{dt}=0 \Rightarrow i = \frac{\mathcal E_{\rm cell}}{L}\,t$

After time $t$ the work done by the cell is $\displaystyle \int ^t_0 \mathcal E_{\rm cell} \,i\, dt = \frac{\mathcal E_{\rm cell}^2 \,t^2}{2L}= \frac 12 L i^2$ and this is minus the work done by the induced emf, $\displaystyle \int ^t_0 \mathcal E_{\rm induced} \,i\, dt=\int ^t_0 \left ( -L \frac {di}{dt}\right) \,i\, dt $, and represents the energy stored by the inductor.

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Negative just signifies emf will such be induced that current induced will oppose change in magnetic field though it.It is not related to work it is just for above reason.

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In faraday's law the negative sign is given as per lenz's law. Lenz's law states that the direction is always such that it will oppose the change in flux which produced it.

This means that any magnetic field produced by an induced current will be in the opposite direction to the change in the original field. E=− dΦ/dt

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    $\begingroup$ I understand that the negative sign is due to Lenz's Law but how does it allow the work done by the induced electric force to be negative if the electric force is not acting against the direction of the current (but instead along with it...seems slightly paradoxical to me)? $\endgroup$
    – Chern-Simons
    Commented Jul 21, 2019 at 7:07

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