Question

A cylindrical conductor has a uniform cross-section. The resistivity of its material increases linearly from left end to right end. If a constant current is flowing through it and at a section distance $x$ from left end, the magnitude of electric field intensity is $E$ , which of the following graphs is correct?

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Given,
Cross-sectional area $\left(A\right)$ of the conductor is constant and
Current $\left(i\right)$ through the conductor is constant.
Since, $A$ and $i$ are constant, the current density $\left(J\right)$ is also constant because, $J=\frac{i}{A}$ .
The relation between the Electric field $\left(E\right)$ and the resistivity $\left(\rho \right)$ of a wire is given by,
$E=\rho J$
$\Rightarrow E \propto \rho $ .
$E=\frac{i \rho }{A}=\frac{i \left(\right. \left(\rho \right)_{0} + a x \left.\right)}{A}$
Therefore,
As the resistivity of its material increases linearly from the left end to the right end, the Electric field also increases linearly with $x$ , i.e., from the left end to the right end.
Therefore, $E-x$ graph will be
.