# Different between spatial change in magnetic field & motional emf?

For the case of a stationary loop, and a changing magnetic field producing a non-conservative electric field $$E_{nc}$$:

If the induced emf ($${\Large{\varepsilon}}$$) is due to both the change in magnetic field strength, and spatial change(due the magnetic field source's motion) an equation to model the sum of two effects is:

$${\Large{\varepsilon}}= \oint E_{nc} \cdot dl=\frac{\partial\Phi_B}{\partial t}=\frac{\partial B}{\partial t}\cdot S\cdot\cos(\alpha)+B\frac{\partial S}{\partial t}\cos(\alpha)$$

How is the final term $${\Large(\small{B\frac{\partial S}{\partial t}\cos(\alpha)}}\Large{)}$$ different than motional emf $$v_xBL$$?

I know in this case the loop is stationary, thus $$v_x$$ = $$0$$.

They seem to be the same mathematically, however, two different causes of effect.

Diagram: • Credit to user @FGSUZ for providing the equation in this question: : physics.stackexchange.com/questions/458890/… – Geodesic Mar 24 '19 at 19:51
• I'll be glad to answer when I understand the question, but I'm sorry, I'm still not getting you. What do you mean? Also, you should include the diagram of your problem. The formula $\Phi_B=B\cdot S\cdot \cos (\alpha)$ is only valid for some easy geometries (the general case needs the integral form). – FGSUZ Mar 24 '19 at 23:18
• Added a diagram. Question simplified: If term 2 is a function of spatial variation(applied to case(2) in the diagram) how is it different from motional emf (in case (1))? If $v = v'$. – Geodesic Mar 24 '19 at 23:40
• Hope that clarifies it. – Geodesic Mar 25 '19 at 0:01

Okay, now I understand your question.

The answer is simple: of course both situations yield the same result, and the same formula.

Then you say that they are different phenomena. Well, I say they are not. Because what matters here is relative motion; that is, the movement of the field with respect to the loop. It is not about "absolute motion with respect to a certain fixed point", but rather, it is about relative displacement between two elements.

Let's go to the simple case: suppose that $$\vec{B}$$ is uniform on its region and it doesn't change in time. Then, the only contribution is due to the change in $$S$$.

The formula $$\Phi_B=B\cdot S$$ can only be applied in some easy geometries like these. The general formula is $$\Phi_B=\iint \vec{B}\cdot d\vec{S}$$, but, as we are considering that $$B$$ is constant, and the angle between both vectors is constant (and also $$0º$$), we can say that $$\Phi_B=B\cdot S$$.

If $$B$$ is constant, then we have that $$\frac{\partial{\Phi_B}}{\partial t}=B\cdot\frac{\partial S}{\partial t}$$

But $$S$$ is the surface of the loop exposed to the magnetic field.

• If the loop moves away from the field, that term will decrease.
• If the magnetic field slowly moves away, that term will decrease as well.

And that's because the formula only cares about how much surface is exposed to the field. It doesn't care wether it's decreasing because it is moving or because the field is moving. Consequently, it's just about the relative motion. The formula only cares about relative motion.

And this is another example of the relativity principle. Explained easily: imagine you're travelling by train, on a perfectly straight way, and with extremely accurately constant velocity. Now you look through the window. You deduce you're moving, because the landscape is variating.

However, how do you know that you're moving forward? It could be the landscape moving backwards! Of course you know the train moves, because you've been taught so. But if no one had told you... if you had been born in the train... you couldn't tell.

This is the same. The loop doesn't know if the loop itself is moving, or if the whole rest of the universe (magnetic field included) is moving backwards.

Fortunately, the loop doesn't care. The physical laws are the same in both considerations.

What's more, you can go further. The loop is moving with respect to you. However, if you moved with the loop, you'd be seeing the magnetic field moving and not the loop.

This is changing from a fixed reference frame, to a moving reference frame. As it is straight line and uniform speed, all laws must be the same. All phenomena you observe should be the same.

Well, in fact it is, you just obtained it.

PS: calling it "motional emf" is not very useful, in my view. It's just "emf".

• Thank you again, for some reason I thought those to cases are totally different. But the more I dive into the formalism regardless of what the "effect" is, the mathematics shows the results being the same regardless of what reference frame this is. Does not matter if it's due to the magnetic force acting on the charges, or the electric force from the non-conservative electric field's curl acting on the charges due the change in $B$, they all lead to the same result so long as the initial & final variables(magnetic flux) are the same for both reference frames. – Geodesic Mar 26 '19 at 12:59
• My diligent review of this topic is to assure I'm not always relying on the flux law for certain cases, because from Feynman & Griffith's book covering electrodynamics, there was a discussion about Faraday's paradox, which is why sometimes focusing on the effects is useful to validate the flux law's application to problem, however, to problems relative to "change in reference" I think it's always applicable. What do you think? – Geodesic Mar 26 '19 at 13:04
• Sorry, I'm not sure I understand what you mean. – FGSUZ Mar 26 '19 at 21:16
• No worries, it was my mistake. As I always do when I have a lot of ideas(some of which are confusing) conflicting with one another. What I tried to say is; The reason I keep reviewing these two cases(diagrams above and their dynamics w.r.t induced $\varepsilon$) is to avoid paradoxical situation such as the ones descried by Faraday's Paradox here: en.wikipedia.org/wiki/Faraday_paradox – Geodesic Mar 27 '19 at 9:18
• And what I noticed from my reviews and your responses, is that the flux law in it's simplicity $\frac{\partial\Phi_B}{\partial t}$ is not always best to use directly. One must analyze the Lorentz force , and Maxwell-Faraday's equation for all cases. Which my examples(the two cases above) qualify perfectly. And beautifully illustrate the two approaches for two dynamics cases. – Geodesic Mar 27 '19 at 9:22

Assuming you have a loop, with the area $$S = W \cdot L$$, like in an dc motor, and $$L$$ keeps constant, then the term

$$B\frac{\partial S}{\partial t}\cos(\alpha)$$

simplifies to

$$B\frac{\partial W}{\partial t}\cdot \cos(\alpha) \cdot L = B \cdot v \cdot \cos(\alpha) \cdot L$$

where $$v$$ is the velocity.

• Great, but the result is not considered to be motional emf correct? Rather induced emf from a non-conservative electric field? – Geodesic Mar 24 '19 at 21:30
• this is the motional emf – abu_bua Mar 24 '19 at 21:41
• How so, if the loop is stationary? – Geodesic Mar 24 '19 at 22:34
• if you mean the dc motor (generator), cos(alpha) changes with time. – abu_bua Mar 24 '19 at 23:09
• Not specifically, could be a motor, could be a stationary loop fixed into place while the magnet(magnetic field source) is moving. So, I believe the second term is a function of spatial changes therefore; $$B\frac{\partial S}{\partial t}\cos(\alpha)$$ can be applied for two cases: 1) Motional emf where the loop is moving w.r.t to a fixed magnet 2) Spatial variance of the magnetic field; as the magnet movies away from the loop. What do you think? – Geodesic Mar 24 '19 at 23:12