Shown in the figure is a semicircular metallic strip that has thickness and resistivity . Its inner radius is and outer radius is . If a voltage is applied between its two ends, a current flows in it. In addition, [2020]

Question

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 (ignoring any role of the magnetic field due to the current). Then (the figure is schematic and not drawn to scale) [2020]

Smart Achievers · Accepted Answer

(1, 3, 4)

Resistance of elementary strips

$\int \frac{1}{d R} = \int_{R_{1}}^{R_{2}} \frac{t d x}{ρ π x}$ $\Rightarrow \frac{1}{R} = \frac{t}{π ρ} \ln (\frac{R_{2}}{R_{1}})$

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

Hence, option (1) is correct.

For the circular motion of electrons, $Δ V$ develops.

The inner surface is at a higher potential, so that the electric field develops radially outward.

Therefore, option (3) is correct.

$\frac{m V_{d}^{2}}{r} = q \vec{E}$ $\Rightarrow \vec{E} = \frac{m V_{d}^{2}}{q r}$

Using

$V_{d} = \frac{I}{n e A},$

we get

$E = \frac{m}{q r} {(\frac{I}{n e A})}^{2} = \frac{m I^{2}}{n^{2} e^{2} A^{2} q r}$

Since

$Δ V = \int \vec{E} \cdot d \vec{r}$ or $Δ V \propto V_{d}^{2}$

$∴$ $Δ V \propto I^{2}$

Therefore, option (4) is correct.

Solution

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

2 The outer surface is at a higher voltage than the inner surface.

3 The outer surface is at a lower voltage than the inner surface.

4 $Δ V \propto I^{2}$

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