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
Best Answer
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.