# Explanation of rotational EMF

Why does an EMF $$\frac{1}{2}B\omega R^2$$ get created across the radius $$R$$ of a Faraday's disc when it rotates, with angular velocity $$\omega$$, in a magnetic field constant with both space and time?

I found about this 'Faraday Paradox' situation on Google. It involves something like: while calculating the motional emf generated, we have to take into consideration the total derivative of the magnetic field and not just the partial derivative. They meant to say that emf generated is negative times the [partial differentiation of magnetic field with time] multiplied by the area + the differentiation of [magnetic field multiplied by area] with time. I am a biology student so can anyone please convert the other equations into words to get the difficulty level down to 80%? Source: Wikipedia

• Did you research already for Faraday disc? – Thomas Fritsch Aug 24 '19 at 13:35
• @Thomas Fritsch No I didn't. I thought it would be better to ask at stackexchange. I am in class 12th and I needed help in understanding this concept. So should I just delete this question and go somewhere else on Google? – user233565 Aug 24 '19 at 13:43
• Okay I found about this 'Faraday Paradox' situation on Google. It involves something like: while calculating the motional emf generated, we have to take into consideration the total derivative of the magnetic field and not just the partial derivative. They meant to say that emf generated is negative times the [partial differentiation of magnetic field with time] multiplied by the area + the differentiation of [magnetic field multiplied by area] with time. I am a biology student so can anyone please convert the other equations into words to get the difficulty level down to 80%? Source:Wikipedia – user233565 Aug 24 '19 at 14:10
• @Thomas Fritsch why should I merge a partial answer into a question? – user233565 Aug 24 '19 at 14:21
• Hey, I did as you said. Thanks for your help. – user233565 Aug 24 '19 at 15:22

Motional emf in rotating rod is nothing but subset of motional.emf ,$$Bvl=V$$ ,where in case of rotating body ,v is replace by $$v=wr ,now$$B.wr.dr integrate it from 0 to R you get EMF $$\frac{1}{2}B\omega R^2$$ ,that is mathmatical view,let us understand it by the point of you of physics,

As you know about motion of conducting rod in uniform magnetic field,produces emf in the body ,as moving charge will experience lorentz ,force and E( Electric field )is create ,due to accumlation of charge ,similarly in case of rotating body ,electron in conducting material moves in circle ,will also experience lorentz force ,and direction of induced emf ,can be understand by flemmings left hand rule. You can understand more about it in the book Concept of physics by hc verma.

As you mention in your comment,that you do not know about faraday law,faraday said that whenever magnetic flux changes ,through closed loop it induces emf ,it is standard and main theory ,where magnetic flux is Φ=BAcosθ And - time of differentiation of it gives emf, Direction of induced emf is such that it ,oppose its own cause ,mean direction of induced current ,magnetic field is such that it decrease incoming flux.

And About faraday paradox i would say that,it is not necessary that all law,always valid for all situation,we should use the laws $$\nabla\times E = -dB/dt$$ and $$F=qv\times B$$, which are universally valid.and basic of all E.M.I in conductors.

• You recalled me of another of my doubt about motional emf. While proving it, we considered the rod to be connected to a conducting rectangular loop (whose one side, say right side, is the rod itself with the top and bottom sides extending to infinity towards right. The derivation uses the fact that as we slide the rod towards right, the magnitude of magnetic flux passing through the area of the loop increases, to oppose which a current sets up in the loop such that the direction of the magnetic field it would produce (inside the loop) due to its flow, will oppose the change in... – user233565 Aug 24 '19 at 14:19
• ...the increased magnetic field. But now how can we extrapolate this derivation to cases in which the loop (the other 3 sides) is entirely missing? If there is no loop, then there is no region through which the flux passing change, upon moving a rod linearly in a magnetic field constant with space and time. Then why? – user233565 Aug 24 '19 at 14:24
• I know it is derived through Lorentz Force equation. In fact I actually, in a theoretical manner, derived it in the above two comments. But you didn't answer my question! How can we extrapolate the result to cases of motion of rod without the loop when we ourselves derived it for the case with the loop? If your answer is still the same then please be kind enough to elaborate (in fact, provide the derivation) – user233565 Aug 24 '19 at 15:05
• Also when did I say I don't know about the Faraday's law? I said I don't know about the 'Faraday Paradox'. It is the case where Faraday's law fails and had been corrected by JC Maxwell in the form of Faraday-Maxwell equations. – user233565 Aug 24 '19 at 15:08
• I don't know what you want to say. Anyway, I already created a new question about this new topic on stackexchange. So this conversation doesn't need to be extended any further over here – user233565 Aug 24 '19 at 15:30

I find notions of flux in this context get confusing. Falling back to the Lorentz force is helpful. For a hypothetical charge on the disk, $$\vec{F}=q\vec{v}\times\vec{B}=q\omega r\hat{\theta} \times B\hat{k}=q\omega B r\hat{r}$$. The work done moving the charge from the center to the rim is $$\int_0^R \vec{F}\cdot\vec{dr}=(1/2)q\omega BR^2$$. EMF is energy per charge so divide by $$q$$ to get the result.

I'm not sure how to bring in area. If we have field $$\vec{B}=B\hat{k}$$ we need an area element of $$rdr\frac{d\theta}{dt}\hat{k}$$ for the flux integral to yield the same result as the integral given by the Lorentz force. $$\frac{dA}{dt}=rdr \frac{d\theta}{dt}$$. So we do recover that the rate of change of the flux equals our EMF, but I'm not sure how to justify consideration for that specific area element.