JEE Advance - Physics (2020 - Paper 1 Offline - No. 10)
Shown in the figure is a semicircular metallic strip that has thickness t and resistivity $$\rho $$. Its inner
radius is R1 and outer radius is R2. If a voltage V0 is applied between its two ends, a current I flows
in it. In addition, it is observed that a transverse voltage $$\Delta $$V 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 (figure is schematic and not drawn to scale)
$$I = {{{V_0}t} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$$
the outer surface is at a higher voltage than the inner surface
the outer surface is at a lower voltage than the inner surface
$$\Delta $$V $$ \propto $$ I2
Maelezo
All the elements are in parallel.
$$ \therefore $$ $$\int {{l \over {dr}} = \int\limits_{{R_1}}^{{R_2}} {{{t\,dx} \over {\rho \,\pi x}}} } $$
$${l \over r} = {t \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$$
Resistance = $${{\pi \rho } \over {t\ln \left( {{{{R_2}} \over {{R_1}}}} \right)}}$$
$$ \therefore $$ $$I = {V \over {{\mathop{\rm Re}\nolimits} sis\tan ce}}$$
$$ = {{{V_t}} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$$
$$eE = {{m{v^2}} \over r}$$
$$I = neA{V_d}$$
$${{{V_0}tdr} \over {\rho \pi r}} = ne(drt)V$$
$$ \Rightarrow V = \sqrt {{{{V_0}} \over {\rho \pi ner}}} $$
$$ \therefore $$ $$E = {m \over {er}}{{V_0^2{t^2}} \over {{\rho ^2}{\pi ^2}{n^2}{e^2}{r^2}}}$$
$$ \Rightarrow E = {{mV_0^2} \over {{e^3}{r^2}{\rho ^2}{\pi ^2}{n^2}}}$$
$$dV = - Edr \Rightarrow dV = - {K \over {{r^3}}}dr$$
$$V = K\int\limits_{{R^1}}^{{R_2}} {{r^{ - 3}}dr} $$
$$\Delta V = {{mV_0^2} \over {{e^3}{\rho ^2}{\pi ^2}{n^2}}}\left( {{1 \over {r_1^2}} - {1 \over {r_2^2}}} \right)$$
$$\Delta V \propto V_0^2 $$
$$ \Rightarrow $$ $$\Delta $$V $$ \propto $$ I2 ( As $$ I \propto {V_0}$$ )