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PhysicsMultiple correct+4 / -2Question 11 of 54

JEE Advanced 2020 · Paper 1, Physics question 11

Shown in the figure is a semicircular metallic strip that has thickness tt and resistivity ρ\rho. Its inner radius is R1R_1 and outer radius is R2R_2. If a voltage V0V_0 is applied between its two ends, a current II flows in it. In addition, it is observed that a transverse voltage ΔV\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)

A semicircular metal strip (shaded annular half-ring) with inner radius R₁ and outer radius R₂ from a common centre. Its two flat ends are connected through wires to a cell V₀; current I enters one end and leaves the other. ΔV is marked across the strip between the inner and outer surfaces.
Choose one or more options
Answer and worked solution

Official answer: A, C, D

(A) Current. Split the strip into semicircular paths of radius rr, width drdr: each has length πr\pi r and cross-section t drt\,dr, and all are in parallel between the ends.
G=∫R1R2t drρπr=tπρln⁡R2R1  ⟹  I=V0tπρln⁡R2R1.G=\int_{R_1}^{R_2}\frac{t\,dr}{\rho\pi r}=\frac{t}{\pi\rho}\ln\frac{R_2}{R_1}\implies I=\frac{V_0t}{\pi\rho}\ln\frac{R_2}{R_1}.
(A) is true.

(B)/(C) Sign of ΔV\Delta V. The electrons drift along circular arcs, so they need a force towards the centre. It comes from a radial electric field: −eE⃗-e\vec E must point inward, so E⃗\vec E points outward. The potential falls outward, so the outer surface is at the lower potential. (C) is true and (B) is false.

(D) The required field is E=mvd2/(er)E=m v_d^2/(er) and vd∝Iv_d\propto I, so ΔV∝I2\Delta V\propto I^2. (D) is true.

(A), (C), (D)

Topics:Ohm's law and resistivityCircular motion dynamicsDrift velocity and current
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