Question

The current in the rod is $(I_o=\frac{V_o}{R_o})$

• A. $I_o{e}^{x/L}$
• B. $\frac{I_o}{L}(L-X)$
• C. $I_o{e}^{-x/L}$
• D. $\frac{I_o}{L}(L+X)$

Answer 1

The correct option is A $I_o{e}^{x/L}$

Consider an element $dx$ at $x=0$.
Let $i\left(x\right)$ be the current through the element. Then,
$\frac{\rho \left(x\right)}{A}dxi\left(x\right)=-dV\left(x\right)$
where, $dV\left(x\right)$ is the potential decreasing from $x=0$ to $x=L$
Hence, $\frac{i\left(x\right)}{A}=-\frac{dV}{dx}.\frac{1}{\rho \left(x\right)}=E\left(x\right).\frac{1}{\rho \left(x\right)}$.....($E\left(x\right)=-\frac{dV}{dx}$)
$i\left(x\right)=\frac{A}{\rho \left(x\right)}.E\left(x\right)$ ............(1)
Now, the potential difference between a point $x=0$ and $x=L$ is given by
$V\left(x\right)=V_0\frac{[L-x]}{L}$
$V\left(x\right)=V_0[1-\frac{x}{L}]$ ...........(2)
Hence, the electric field produced in rod at any point, $0\le x\le L$ is
$E\left(x\right)=-\frac{\partial V\left(x\right)}{\partial x}$
$E\left(x\right)=-V_0[0-\frac{1}{L}]$ ............from(2)
$\therefore E\left(x\right)=\frac{V_0}{L}$
using this value in equation(1) we
$i\left(x\right)=\frac{A{V}_0}{L\rho \left(x\right)}$
$i\left(x\right)=\frac{A{V}_0}{L}\frac{1}{{\rho }_0}{e}^{x/L}$
$i\left(x\right)=\frac{V_0}{R_0}.e^{x/L}$
Put, $I_0=\frac{V_0}{R_0}$
$\therefore i\left(x\right)=I_0.e^{x/L}$