Question

Consider a wire ribbon made from a 2-dimensional conductor with reciprocal mass tensor with

components bxx=3, byy=4, bxy=byx=0 in units of inverse electron mass. Assume the wire extends mainly
in the x-direction, but with some width in the y-direction. Assume the collision time is independent of
direction and has a value of 2.7E-11 sec. Assume two electrons per atom with a square lattice of spacing
0.1 nm.

What is the current density (magnitude and direction) when an electric field of 1 V/m is applied
along the +x direction?

Answer 1

To find the current density for a 2-dimensional conductor under the given conditions, we use the Drude model and Ohm's Law in conjunction with the provided mass tensor components, collision time, electron density, and lattice spacing.

Step-by-step explanation:

To find the current density when an electric field of 1 V/m is applied along the +x direction in a 2-dimensional conductor with the given mass tensor components and collision time, we use the Drude model of conductivity. First, we calculate the conductivity tensor components (σxx and σyy) using the reciprocal mass tensor components (bxx and byy) and the given collision time (τ).

Using the formula σxx = ne²τbxx and σyy = ne²τbyy, where e is the electron charge and n is the electron density per volume, we can then compute the current density using Ohm's Law in tensor form, J = σE, where E is the electric field.

To find n, we use the fact that there are two electrons per atom in a square lattice of 0.1 nm spacing, thus n = (2 atoms / unit cell)/(0.1 nm) ³. Finally, with n and e known constants, and the electric field given, we can calculate the current density magnitude and direction.