## Drude model with magnetic field The setup is shown in the sketch below. ![[Drawing 2024-09-03 23.48.34.excalidraw.svg]] $\mathbf{E} = (E_x, 0, 0), \quad \mathbf{B} = (0, 0, B_z).$ Drift velocity is at $y$ direction. So Lorentz force would be $\mathbf{F} = -e \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right)$ We may get cyclotron frequency $\omega_c = \frac{|e| B_z}{m^*}$ and drift velocity $v_D = \frac{E_x}{B_z}$ $\mathbf{v}(t) = \begin{pmatrix} 0 \\ -v_0 \end{pmatrix} - \begin{pmatrix} v_y(0) + v_0 \\ -v_x(0) \end{pmatrix} \sin(\omega_c t) + \begin{pmatrix} v_x(0) \\ v_y(0) + v_0 \end{pmatrix} \cos(\omega_c t)$ This motion is called $\mathbf{E\times B}$ drift. >[!Notice] >We have assumptions: > - scattering for angular dependence is assumed to be nonexistent. > - scattering time follows Poisson distribution. For $\omega_{c} \tau \ll1$, or say under small magnetic field (namely, Hall effect condition), the carrier cannot complete a cyclotron orbit without being scattered. For $\omega_{c}\tau \gg1$, or say under large magnetic field (namely [[Integer quantum Hall effect|QHE]] or [[Magnetotransport in 2D and Shubnikov–de Haas effect|Shubnikov-de Haas effect]] condition), no scattering is possible. >[!Note] >The time difference between QHE and Shubnikov-de Haas effect is the quantum lifetime, $\omega_{c} \cdot \tau_{q}$. If consider the collision process, we may define the mobility $\mu$, $\mu = \frac{|e| \tau}{m^*}$, and the Hall angle, $\theta$, $\tan \theta = \frac{\bar v_y}{\bar v_x} = \omega_c \tau = \mu B$. Here this $B$ is the equivalent magnetic field at $z$ direction, and the expression is valid for general field. $\bar v$ stands for the statistical result (Poisson distribution), check the [[Semiconductor Nanostructures|book]] chapter 10.3 for the exact expression. And with the Hall bar geometry (see later sketch), we will have $\sigma_{xx}(B) = \frac{n_s e^2 \tau}{m^*} \cdot \frac{1}{1 + \omega_c^2 \tau^2}$ $\sigma_{xy}(B) = \frac{n_s e^2 \tau}{m^*} \cdot \frac{\omega_c \tau}{1 + \omega_c^2 \tau^2} = \omega_c \tau \sigma_{xx} = \frac{n_s |e|}{B} - \frac{\sigma_{xx}}{\omega_c \tau}$ With the result, the specific resistivity is $\rho_{xx} = \frac{m^*}{n_s e^2 \tau} = \frac{1}{n_s e \mu}$ $\rho_{xy} = \frac{B}{|e| n_s}$ From $\rho_{xy}$ we have $n_{s}$, which could be read from the slope, as mentioned in [[Metal electrodes on semiconductor surfaces]]. The profile looks like the following, ![[Drawing 2024-09-04 00.36.00.excalidraw.svg]] When $\mu B=1$, $\rho_{xy}=\rho_{xx}$, the $\rho=\frac{1}{n_{s}e\mu}$. This can be directly calculated from $\rho_{xx}$ since $\mu=\frac{|e|\tau}{m^{*}}$. Therefore, we get $n_{s}$ and $\mu$ as, $n_s = \frac{1}{\left| e \right| \left. \frac{d \rho_{xy}}{dB} \right|_{B=0}}$ $\mu = \frac{\left. \frac{d \rho_{xy}}{dB} \right|_{B=0}}{\rho_{xx}(B=0)} = \frac{\frac{1}{\left| e \right| n_s}}{\frac{m^*}{n_s e^2 \tau}} = \frac{\left| e \right| \tau}{m^*}$ The measurement is done on a Hall bar geometry, as shown below. ![[Drawing 2024-09-04 00.49.59.excalidraw.svg]] This geometry is good cause current are forced to flow parallel to the bar. The design also follows two rules: - The separation of a voltage probe and the current contact should be larger than $4W$ to let the Hall angle go fully developed. - To avoid disturbance of the potential, the width of the voltage probes should be much smaller than $W$. Typical size of hall bar: 100 $\rm \mu m$ wide and 1 mm long. And measurement is given by $\rho_{xx} = \frac{U}{I} \cdot \frac{W}{L}, \quad \rho_{xy} = \frac{U_H}{I}$