Question 713: Current Electricity

Question

Shown in the figure is a semicircular metallic strip that has thickness t and resistivity $$\rho $$. Its inner
radius is R
1
and outer radius is R
2
. If a voltage V
0
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)

Accepted Answer

All the elements are in parallel.
$$ 	herefore $$ $$\int {{l \over {dr}} = \int\limits_{{R_1}}^{{R_2}} {{{t\,dx} \over {ho \,\pi x}}} } $$
$${l \over r} = {t \over {\pi ho }}\ln \left( {{{{R_2}} \over {{R_1}}}} ight)$$
Resistance = $${{\pi ho } \over {t\ln \left( {{{{R_2}} \over {{R_1}}}} ight)}}$$
$$ 	herefore $$ $$I = {V \over {{\mathop{m Re}
olimits} sis	an ce}}$$
$$ = {{{V_t}} \over {\pi ho }}\ln \left( {{{{R_2}} \over {{R_1}}}} ight)$$
$$eE = {{m{v^2}} \over r}$$
$$I = neA{V_d}$$
$${{{V_0}tdr} \over {ho \pi r}} = ne(drt)V$$
$$ \Rightarrow V = \sqrt {{{{V_0}} \over {ho \pi ner}}} $$
$$ 	herefore $$ $$E = {m \over {er}}{{V_0^2{t^2}} \over {{ho ^2}{\pi ^2}{n^2}{e^2}{r^2}}}$$
$$ \Rightarrow E = {{mV_0^2} \over {{e^3}{r^2}{ho ^2}{\pi ^2}{n^2}}}$$
$$dV =  - Edr \Rightarrow dV =  - {K \over {{r^3}}}dr$$
$$V = K\int\limits_{{R^1}}^{{R_2}} {{r^{ - 3}}dr} $$
$$\Delta V = {{mV_0^2} \over {{e^3}{ho ^2}{\pi ^2}{n^2}}}\left( {{1 \over {r_1^2}} - {1 \over {r_2^2}}} ight)$$
$$\Delta V \propto V_0^2 $$
$$ \Rightarrow $$ $$\Delta $$V $$ \propto $$ I
2
( As $$ I \propto {V_0}$$ )

Answer

$$I = {{{V_0}t} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$$

Answer

the outer surface is at a higher voltage than the inner surface

Answer

the outer surface is at a lower voltage than the inner surface

Answer

$$\Delta $$V $$ \propto $$ I
2

Current Electricity

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