Two wires have a nearly equal electron number density. Assume that the electric field in each wire points directly along the wire, that electric fields in the two wires have the same magnitude, but that the conductivity in wire $A$ is twice that in wire $B$. This means that the drift velocity in wire $A$ is $b$ times that in wire $B$, where $b$ is
A) 4 .
B) 2.
C) 1 .
D) $1 / 2$.
E) $1 / 4$.
F) Impossible to determine from the given information.

Question

Quizplus · Accepted Answer

The drift velocity ($v_d$) in a conductor is directly proportional to the electric field ($E$) and the conductivity ($\sigma$) of the material, as given by the relation $v_d = \frac{E}{\rho}$ or $v_d = \sigma E$ where $\rho$ is the resistivity. Since the conductivity of wire $A$ is twice that of wire $B$ and the electric fields have the same magnitude, the drift velocity in wire $A$ will be twice that in wire $B$.

Two Wires Have a Nearly Equal Electron Number Density AAA Is Twice That in Wire

Two Wires Have a Nearly Equal Electron Number Density $A$ Is Twice That in Wire