Shown in the figure is a semicircular metallic strip that has...

Resistance of the strip $= ρ\dfrac{\pir}{tdr}$

$R =$net resistance

$\dfrac{1}{R} = ∫ \dfrac{tdr}{ρ\pir}\dfrac{t}{\piρ}{∫}_{{R}_{1}}^{{R}_{2}}\dfrac{dr}{r}$

$R = \dfrac{\piρ}{tln\left(\dfrac{{R}_{2}}{{R}_{1}}\right)}$

Hence, current i $= \dfrac{{V}_{0}}{R} = \dfrac{{V}_{0}t}{\piρ}\ln \left(\dfrac{{R}_{2}}{{R}_{1}}\right)$

for elementary dr,

$di = \dfrac{{V}_{0}}{ρ\left(\dfrac{\pir}{tdr}\right)} = \dfrac{{V}_{0}tdr}{ρ\pir}$

$i = nAev$

$\dfrac{{V}_{0}tdr}{ρ\pir} = nt\left(dr\right)ev$

$eE = m\dfrac{{v}^{2}}{r} = m{V}_{0}^{2}{n}^{2}{ℓ}^{2}{ρ}^{2}{\pi}^{2}{r}^{2} \times \dfrac{1}{r}$

$m\dfrac{{v}^{2}}{r} = m{V}_{0}^{2}{n}^{2}{ℓ}^{2}{ρ}^{2}{\pi}^{2}{r}^{2} \times \dfrac{1}{r}$

$E ∝ \dfrac{1}{{r}^{3}}$

$E = k\dfrac{1}{{r}^{3}}$

$\DeltaV = ∫ Edr = k{∫}_{{R}_{1}}^{{R}_{2}}\dfrac{dr}{{r}^{2}} ∝ {V}_{0}^{2}$

$∴ ∝ {i}^{2}$