Q: Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ . Its inner radius is R1 and outer radius R2 . If a voltage Vo is applied between its two ends , a current I flows in it . In addition , it is observed that a transverse voltage ΔV develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current ) . Then

(a) $I = \frac{V_o}{\pi \rho} ln(\frac{R_2}{R_1})$

(b) the outer surface is at a higher voltage than the inner surface

(c) the outer surface is at a lower voltage than the inner surface

(d) $\Delta V \propto I^2 $

**Click to See Solution : **

$\int \frac{1}{dr} = \int_{R_1}^{R_2} \frac{t dx}{\rho \pi x}$

$\frac{1}{r} = \frac{t}{\pi \rho}ln(\frac{R_2}{R_1}) $

$ Resistance = \frac{\pi \rho}{t ln(\frac{R_2}{R_1})} $

$I = \frac{V}{Resistance} $

$I = \frac{V t}{\pi \rho}ln(\frac{R_2}{R_1}) $

$eE = \frac{m v^2}{r}$

$I = n e A v_d$

$\frac{V_o t dr}{\rho \pi r} = n e (dr t)v $

$v = \frac{V_o }{\rho \pi n e r}$

$E = \frac{m}{e r} \frac{V_o^2}{\rho^2 \pi^2 n^2 e^2 r^2}$

$E = \frac{m V_o^2}{e^3 r^2 \rho^2\pi^2 n^2} $

$dV = -E dr $

$dV = – \frac{K}{r^3}dr $

$V = K \int_{R_1}^{R_2}r^{-3} dr $

$\Delta V = \frac{m V_o^2}{e^3 r^2 \rho^2\pi^2 n^2} (\frac{1}{r_1^2} – \frac{1}{r_2^2})$

$\Delta V \propto V_o^2 $

$I \propto V_o $

$\Delta V \propto I^2 $