JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Current Electricity

Shown in the figure is a semicircular metallic strip that has thickness t and resistivity

ρ

. 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

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)-

(A) $I = \frac{V _{0} t}{π ρ} ln (\frac{R _{2}}{R _{1}})$ ✓ Correct
(B)the outer surface is at a higher voltage than the inner surface
(C)the outer surface is at a lower voltage than the inner surface✓ Correct
(D) $Δ V \propto I^{2}$ ✓ Correct

Answer: (A), (C), (D)

Solution

Consider a thin radial element at distance x with thickness dx. Its resistance dR = ρdx/(t·πx). Total resistance: 1/R = ∫(t dx)/(ρπx) from R₁ to R₂ = (t/πρ)ln(R₂/R₁), so R = πρ/[t ln(R₂/R₁)]. Current I = V₀/R = V₀t ln(R₂/R₁)/(πρ) ✓. For circular motion, electrons need centripetal force provided by radial electric field: mv_d²/r = qE, where E points radially outward. This makes the outer surface at lower potential ✓. Since E ∝ v_d² and v_d ∝ I, we have ΔV ∝ I² ✓.

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Current Electricity

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