Replace 3_drude_model_solutions.md

Added / expanded solutions to the exercises

docs/3_drude_model_solutions.md

@@ -13,24 +13,23 @@ $$
 V_H = -\int_{0}^{W} E_ydy
 $$
 
-where $E_y = -v_xB$.
-
-$R_{xy}$ = $-\frac{B}{ne}$, so it does not depend on the sample geometry.
+where $E_y = v_xB$, which we obtain from the steady-state Drude equation. The (surface) current density is $I = j_x W$. Thus $R_{xy} = \frac{v_xB}{j_x} = \frac{-B}{ne}$,
+so it does not depend on the sample geometry.
 
 
 #### Question 2.
 If hall resistance and magnetic field are known, the charge density is calculated from $R_{xy} = -\frac{B}{ne}$.
 
-As $V_x = -\frac{I_x}{ne}B$, a stronger field makes Hall voltages easier to measure.
+As $V_x = -\frac{I_x}{ne}B$, a stronger field makes Hall voltages easier to measure. A lower charge density does not mean that the Hall voltage is easier to measure, since the current depends also on the charge density.
 
 #### Question 3.
-
+Using that $ I = j_x W$, $E= LE $ and $E = \rho_{xx} j_x$ we obtain:
 
 $$
 R_{xx} = \frac{\rho_{xx}L}{W}
 $$
 
-where $\rho_{xx} = \frac{m_e}{ne^2\tau}$. Therefore, scattering time ($\tau$) is known and $R_{xx}$ depend upon the sample geometry.
+where $\rho_{xx} = \frac{m_e}{ne^2\tau}$. Therefore, scattering time ($\tau$) is known and $R_{xx}$ depend upon the sample geometry, but $\tau$, $n$ do not.
 
 <!---
 #### Question 4.
@@ -47,10 +46,11 @@ See 3_drude_model.md
 Find electron density from $n_e =   \frac{ZnN_A}{W} $
 
 where *Z* is valence of copper atom, *n* is density, $N_A$ is Avogadro constant and *W* is atomic weight. Use $\rho$ from the lecture notes to calculate scattering time.
+($ \tau = 2.57 \cdot 10^{-14} s$)
 
 #### Question 2.
 $\lambda = \langle v \rangle\tau$
-
+($\lambda =3 nm $)
 #### Question 3.
 Scattering time $\tau \propto \frac{1}{\sqrt{T}}$; $\rho \propto \sqrt{T}$
 
@@ -63,20 +63,47 @@ In general, $\rho \propto T$ as the phonons in the system scales linearly with T
 $\rho_{xx}$ is independent of B and $\rho_{xy} \propto B$
 
 #### Question 2.
-4 (Refer to the lecture notes).
-
-#### Question 3.
 
 $$
-\sigma_{xx} = \frac{\rho_{xx}}{\rho_{xx}^2 + \rho_{xy}^2}
+\sigma_{xx} = \frac{\rho_{xx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{mne^2\tau}{m^2+e^2\tau^2 B^2}
 $$
 
 $$
-\sigma_{xy} = \frac{-\rho_{yx}}{\rho_{xx}^2 + \rho_{xy}^2}
+\sigma_{xy} = \frac{-\rho_{yx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{Bne^3 \tau^2}{m^2+e^2 \tau^2 B^2}
 $$
 
+#### Question 3.
 This describes a [Lorentzian](https://en.wikipedia.org/wiki/Spectral_line_shape#Lorentzian).
 
+
+#### Question 4. 
+see lecture notes. The sign of the hall coefficient indicates the dominant charge carriers.
+
+
+### Exercise 4. Positve and negative charge carriers
+
+#### Question 1
+$$ 
+$\bf J$ = -n_e e \mathbf{ v_e} + n_h e \mathbf{ v_h}
+$$
+
+#### Question 2
+Write down the Drude equation for the positive and negative charge carriers seperately, $ \mu_e = \frac{e\tau_e}{m_e}$ and $ \mu_h = \frac{e\tau_h}{m_h}$
+
+#### Question 3
+From the assumption, we can ignore the $(\mathbf{ v_i} \times \mathbf{B})_x$ terms. Then combining question 1 and 2 gives the result. 
+
+
+#### Question 4
+The net current in the y-direction should be zero.
+
+#### Question 5
+$$
+R_H = \frac{n_h\mu_h^2 - n_e\mu_e^2}{e(n_h\mu_h + n_e \mu_e)^2}
+$$
+The sign is determined by a weighted avergage of the charge density and the mobilty of the positive and negative charge carriers.
+<!---
+
 Solution extra exercise:
 
 
@@ -153,4 +180,4 @@ $$
 
 This represents a [cycloid](https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif): a circular motion around a point that moves in the $y$-direction with velocity $v_d=\frac{E_x}{B_z}$.
 
-
+-->