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. The value of current $I$ is given by :

• A. $I=\frac{V_{0}t}{\pi \rho }\ln \left(\frac{R_1}{R_2}\right)$
• B. $I=\frac{V_{0}t}{\pi \rho }\ln \left(\frac{R_2}{R_1}\right)$
• C. $I=\frac{V_0{R}_1{R}_2}{\pi \rho t}\ln \left(\frac{R_1}{R_2}\right)$
• D. $I=\frac{\pi V_{0}t}{\rho }\ln \left(\frac{R_1}{R_2}\right)$

Answer 1

The correct option is B $I=\frac{V_{0}t}{\pi \rho }\ln \left(\frac{R_2}{R_1}\right)$

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

Now ,

All such elements are in parallel

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

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

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

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

So , the option $\left(b\right)$ is correct .