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) [JEE Adv 2020 P1]

{I}={{V}_{0} {t}}/{\\pi \\rho } \ln ({{R}_{2}}/{ {R}_{1}} )

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

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

Current Electricity

Section: Physics

Question : 7 of 29

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Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ. 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 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 (figure is schematic and not drawn to scale)

[JEE Adv 2020 P1]

I=
V0t
πρ
‌ln(
R2
R1
)
The outer surface is at a higher voltage than the inner surface
The outer surface is at a lower voltage than the inner surface
ΔV∝l2

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