Can You Resist The Donut?

Let's take a small volume as shown.

Since in that sector, infinitesimally small resistors are connected in series, $dR = \dfrac{\rho dr}{dS}$

where $dS$ is the cross sectional area. $dS = r \,d\varphi \cdot h$ , so the value becomes $dR = \dfrac{\rho dr}{rh\,d\varphi}\\R = \dfrac{\rho}{h}\int_{r_1}^{r_2} \dfrac 1{r}dr = \dfrac{\rho}{h\,d\varphi}\ln\left(\dfrac{r_2}{r_1}\right)$

Now, we see that there are infinitesimally small sectors arranged parallely. Using $\displaystyle \dfrac 1 R = \sum \dfrac 1 {R_i} = \int \dfrac 1 {R(x)}dx$ , we get,

$d\left(\dfrac 1R\right) = \dfrac{h\,d\varphi}{\rho\ln\left(\dfrac{r_2}{r_1}\right)} \\\dfrac 1 R = \dfrac{h}{\rho_0\ln\left(\dfrac{r_2}{r_1}\right)}\int_0^{2\pi} \dfrac1{|\sec\varphi |}d\varphi = \dfrac{4h}{\rho_0\ln\left(\dfrac{r_2}{r_1}\right)}\\\Rightarrow R = \dfrac{\rho_0}{4h}\ln\left(\dfrac{r_2}{r_1}\right)$

Substituting values, we get $\color{#3D99F6}{\boxed{R = \dfrac{500}3\si{\ohm}}}$

The inner and outer rings are equipotentials, and $\nabla^2\Phi = 0$ , and hence we have $\Phi \; = \; \alpha + \beta \ln r$ for some constants $\alpha,\beta$ . Thus the potential difference is $V = \Delta \Phi =\beta \ln\tfrac{r_2}{r_1}$ .

The current density is $\mathbf{j} \; = \; \rho^{-1}\mathbf{E} \; = \; \rho^{-1} \nabla \Phi \; = \; \frac{|\cos\phi|}{\rho_0} \frac{d \Phi}{dr}\hat{\mathbf{r}} \; = \; \frac{\beta |\cos\phi|}{\rho_o r}\hat{\mathbf{r}}$ and so the current flowing in the circuit is $I \; = \; \int_0^{2\pi} \frac{\beta |\cos\phi|}{\rho_0 r} rh\,d\phi \; = \; \frac{4\beta h}{\rho_0}$ so that the resistance of the annulus is $\frac{V}{I} \; = \; \frac{\rho_0}{4h}\ln\tfrac{r_2}{r_1} \; = \; \frac{10 \times 2}{4 \times 0.03} \; = \; \boxed{166.667} \; \Omega$