Question

Shown in the figure is a semicircular metallic strip that has thickness $t$ and resistivity $\rho$ . Its inner radius is $R_1$ and outer radius is $R_2$ . If a voltage $V_0$ is applied across its two ends, a current $I$ flows in it. In addition, it is observed that a transverse voltage $\Delta V$ develops across 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. $I=\frac{V_{0}t}{\pi \rho }\ln \left(\frac{R_1}{R_2}\right)$
• 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. $\Delta V\propto I^2$

Answer 1

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

$\Delta V\propto I^2$

Let us choose an elementary ring at a distance $x$ from the centre and having thickness $dx$

Now ,

All such elements are in parallel

$\therefore \int \frac{1}{dr}={\int }_{R_1}^{R_2}\frac{tdx}{\rho \pi x}$

$\frac{1}{r}=\frac{t}{\pi \rho }\ln \left(\frac{R_1}{R_2}\right)$

Resistance $=\frac{\pi \rho }{t\ln \left(\frac{R_1}{R_2}\right)}$

Current
$I=\frac{V_{0}t\ln \left(\frac{R_1}{R_2}\right)}{\pi \rho }$

So , the option (a) is correct .

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

So , option (c) is correct .

$\frac{m{V}_d^2}{r}=q\overrightarrow{E}$
$E=\frac{m{V}_d^2}{qr}$ $(I=nA{V}_d\Rightarrow V_d\propto i)$
$\Delta V=\int \overrightarrow{E}.\overrightarrow{dr}$

$\Delta V\propto V_d^2$

$\Delta V\propto I^2$