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)-
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a
I=V0tπρlnR2R1
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
d
ΔV∝I2
answer is A.
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Detailed Solution
All the elements are in parallell=πx and A=tdx⇒dx=ρπxtdx∴∫1dr=∫R1R2 tdxρπx1r=tπρlnR2R1 Resistance =πρtℓnR2R1i=V0tlnR2R1πρ (A)(−eE→) will be inward direction in order to provide centripetal acceleration. Therefore electric field will be radially outwardVouter
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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)-