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 between its two ends, a current I flows in it. In addition, it is observed that a transverse voltage $\Delta V$ 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. $I=\frac{v_{0}t}{\pi \rho }\ln \left(\frac{R_2}{R_1}\right)$
• 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)

Area of the strip $=(t$.dr $)$
$dR=\rho \left(\frac{\pi r}{tdr}\right)$
$\sum \frac{1}{dR}=\underset{R_1}{\overset{R_2}{\int }}\frac{tdr}{\pi \rho r}$
$\frac{1}{R_{eq}}=\left(\frac{t}{\pi \rho }\right)L_n\left(\frac{R_2}{R_1}\right)$
$R_{eq}=\frac{\pi \rho }{t\times \ell n\left(\frac{R_2}{R_1}\right)}$
So, $I=\frac{V_0}{R_{\text{eq }}}=\left(\frac{V_{0}t}{\pi \rho }\right)\ln \left(\frac{R_2}{R_1}\right)$
So, option (A) is correct.

Electrons are revolving in a circular path so there is a centripetal force on the electron, which provide a transverse electric field.
The direction of transverse electric field is radially out side hence inner surface is at higher potential.
So, option (C) is correct.

Current density $J(r)=\frac{dl}{(tdr)}$
$J(r)=\frac{V_0}{(dr)(tdr)}$
$J(r)=\frac{V_0}{\left(\frac{\rho \pi r}{tdr}\right)(tdr)}$
$J(r)=\frac{V_0}{(\rho \pi r)}$
$I=neA{V}_d$
$J(r)=ne{V}_d$
$\Rightarrow V_d=\left(\frac{J(r)}{ne}\right)=\left(\frac{V_0}{\rho \pi rne}\right)$ ....(i)
Centripetal force
$e{E}_T=\frac{m_e{v}_d^2}{r}$ ... (ii)
From (i) and (ii) $E_T\propto V_0^2\dots$ (iii)
$\Delta V=\int E_{T}dr$
$\Rightarrow \Delta V\propto E_T$ .... (iv)
From (iii) and (iv)
$\Delta V\propto V_0^2$
$l\propto V_0$
Hence
$\Delta V\propto l^2$
Hence Option (D) is correct.