How to find the force on a charge that is distanced from a bounded circle with a changing magnetic field?

Question

I've come across the following question, and when I attempted to solve it using different two different methods, I didn't get the same answer. The question presents an area of a circle, with radius $R$, in which there is a changing magnetic field, $\overrightarrow{B}(t)=(\alpha t^2 +\beta t)(-\hat{z})$ . There is a charge, $q$, that is $2R$ from the center of the aforementioned circle. What is the force on the charge?

Method 1: Using Faraday's Law - the "Integrative" way: According to the law: $$\epsilon = -\frac{d}{dt}\Phi$$ First, the flux through the circle of radius $2R$ is the flux through the circle of radius $R$: $$\Phi = \pi R^2(\alpha t^2 +\beta t)$$ $$\epsilon = -\pi R^2(2\alpha t +\beta)$$ And now, using the relation between potential and electric field: $$-\oint Edl= -\pi R^2(2\alpha t +\beta)$$ $$E4\pi R=\pi R^2(2\alpha t +\beta)$$ $$\vec{E}=\frac{R(2\alpha t +\beta)}{4}\hat{\phi}$$ And so the force is: $$\vec{F}=q\frac{R(2\alpha t +\beta)}{4}\hat{\phi}$$ Method 2: Maxwell's formula: $$\nabla \times \vec{E} = -\frac{\partial{\vec{B}}}{\partial{t}} $$ And so, after eliminating the parts of the curl that aren't contributing due to symmetry or the direction: $$\frac{1}{r}(\frac{d}{dr}rE_{\phi})(\hat{z}) = (2\alpha t +\beta)\hat{z}$$ And eventually: $$\vec{E}=\frac{r(2\alpha t +\beta)}{2}\hat{\phi}$$ Substituting $r$ with $2R$, we get: $$\vec{E}=R(2\alpha t +\beta)\hat{\phi}$$ $$\vec{F}=qR(2\alpha t +\beta)\hat{\phi}$$

I would appreciate if someone could elaborate on why I get different answers, and what changes should be done in order to get to the same solution.

Answer 1

there is a mistake in the second procedure, when you try to solve the differential equation. In practice, your equation should be written as: $$ \frac{1}{r} \frac{\partial}{\partial r} r E_{\phi} = \left( 2 \alpha t + \beta \right) \; if \; r < R \\ \frac{1}{r} \frac{\partial}{\partial r} r E_{\phi} = 0 \; if \; r > R.$$ Then, you procede by simple integration. The solution becomes $$E_{\phi}\left( R\right) = \frac{R}{2} \left( \alpha t + \beta \right),$$ and $$E_\phi(2R) = \frac{E_\phi(R)}{2} = \frac{R}{4} \left( \alpha t + \beta \right).$$ As you can see the two solutions are the same.