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 $Δ V$ 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)-

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

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

$I = \frac{V_{0} t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$

$ΔV \propto I^{2}$

answer is A, C, D.

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Detailed Solution

All the elements are in parallel

$l = πx and A = tdx$

$\Rightarrow dx = \frac{ρπx}{tdx}$

$\begin{array}{l} ∴ \int \frac{1}{dr} = \int_{R_{1}}^{R_{2}} \frac{tdx}{ρ π x} \\ \frac{1}{r} = \frac{t}{π ρ} \ln (\frac{R_{2}}{R_{1}}) \\ Resistance = \frac{π ρ}{t ℓ n (\frac{R_{2}}{R_{1}})} \end{array}$

$i = \frac{V_{0} tln (\frac{R_{2}}{R_{1}})}{π ρ} (A)$

$(- e \vec{E})$ will be inward direction in order to provide centripetal acceleration. Therefore electric field will be radially outward

$\begin{array}{l} V_{outer} < V_{inner} (C) \\ \frac{{mV}_{d}^{2}}{r} = q \vec{E} \\ E = \frac{{mV}_{d}^{2}}{qr} (I = ne {AV}_{d} \Rightarrow V_{d} \propto i) \\ ΔV = \int \vec{E} \cdot \vec{dr} \\ ΔV \propto V_{d}^{2} \\ ΔV \propto I^{2} \end{array}$

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