Shown in the figure is a semicircular metallic strip that has thickness t and resistivity . 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 Δ 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 = V₀ t/π ≤ft( 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

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JEE Advanced 2020 Paper 1

Section: Physics

Question : 11 of 54

Marks: +1, -0

t

\rho

R_{1}

R_{2}

V_{0}

\Delta V

Solution:

Elementary resistance

dR = \frac{\rho \times \pi x}{t dx}

\therefore \frac{1}{R} = \int\limits_{R_1}^{R_2} \frac{t dx}{\rho \times \pi x} = \frac{t}{\pi \rho} \ln \left( \frac{R_2}{R_1} \right)

\therefore l = \frac{V_0}{R} = \frac{v_0 t}{\pi \rho} \ln \left( \frac{R_2}{R_1} \right)

And, for circular motion of electron, ΔV develops.

∴ Inner surface at higher potential so that electric field develops radially outward.

\frac{m v^2}{r} = e E

\Rightarrow E = \frac{m}{er} \times \left( \frac{l}{Ane} \right)^2

\therefore \Delta V = \int E dr

\Rightarrow \Delta V \propto l^2

\therefore

Answer

(A, C, D)

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