Question

A metal rod of radius $a$ is concentric with a metal cylindrical shell of radius $b$ and length $l$. The space between rod and cylinder is tightly packed with a high resistance material of resistively $\rho$. A battery having a terminal voltage $V$ is connected across the shown. Neglect resistance of rod and cylinder. If $I$ is the total current in the circuit then:

• A. $I=\frac{lV}{\rho }$
• B. $I=\frac{2\pi lV}{\rho (1nb-1na)}$
• C. $I=\frac{4\pi lV}{\rho (1nb-1na)}$
• D. $I=\frac{lV}{4\pi \rho (1nb-1na)}$

Answer 1

The correct option is B $I=\frac{2\pi lV}{\rho (1nb-1na)}$

$6$ metal rod if radius $a$ is concentric with a

metal cylindrical shall of radius $=b$

Length$=l$

The space between rod and cylinder is tightly packet with high resistance material

$\rho =$ resistivity

a battery having a terminal voltage $=V$

$I=$ total current in the circuit

Let, radial flowest charge between rod and cylinder we have at $\rho$

$J=\frac{I}{2\pi rl}$and

$E=\rho I=\frac{\rho l}{2\pi rl}$

With both J and E in the direction of $\alpha$. Then the definition of the potential.

$dV=-Eds-Edr=-\frac{\rho l}{2\pi l}\times \frac{ds}{V}$

And so nothing the polarity of $V_f$

$V_f={\int }_{r_1}^{r_2}dV=-\frac{\rho V}{2\pi l}{\int }_{\alpha }^b\frac{dr}{V}=\frac{-\rho I}{2\pi l}ln\left(\frac{b}{\alpha }\right)$

Solving for I

$I=\frac{2\pi lV}{\rho (lnb-ln\alpha )}$