Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Physics of Engineered and Nanostructured Materials (SS2011) Sheet5 - 3.6.2011 Drude model and Magneto-Transport a) In the framework of the Drude model, derive the motion equation of an electron of ~ and magnetic field B, ~ under the assumption charge −e as a function of an electric field E of a constant relaxation time τ . In the stationary state and for an electron density n, find ~ and the current density J, ~ for B ~ applied in the relationship between the electric field E ~ the z-direction: B = (0,0,B). What are the resistance and conductivity tensors ρb and σ b, respectively? b) What is the geometrical interpretation of the Hall angle θ defined by tan θ = µB, where µ is the sample mobility? c) In the case of a metallic sample of width w, length l and thickness d, with an electron density n, calculate in the stationary state the value and sign of the Hall coefficient AH for the case of a magnetic field applied in the z-direction and an externally applied electric field Ex in the x-direction, again in the framework of the Drude model. The Hall coefficient is related to the Hall resistivity RH (B) by: RH (B) = AH B d Solution: a) Equation of motion: ~ − m~v + q~v × B ~ , m~v˙ = q E τ Stationary case: ~v˙ = 0 Current density: ~j = −en~v Conductivity: σ0 = neµ q = −e Mobility: µ = eτ /m Using the equations above for the current density, the conductivity and the mobility, the stationary case of the equ. of motion results to: ~ × ~j) ~ = 1 (~j − µB E σ0 Now we can use Ohm’s law ~ = ρb~j E to obtain the resistivity tensor 1 +µB 0 1 −µB 1 0 ⇒ ρb = σ0 0 0 1 conductivity: ρb−1 = σ b σ b= 1 1+µ2 B 2 µB σ0 1+µ 2B2 0 µB − 1+µ 2B2 1 1+µ2 B 2 0 0 0 1 ~ and ~j not parallel (angle θ = tan µB between them). As shown in b) Ohm’s law ⇒ E ~x the figure below, the Hall angle is defined as the angle between the external electric field E ~ ~ ~ ~ and the total field E = Ex + Ey . Here, Ey origins from the deflection of the electrons in the magnetic field due to the Lorentz force. j0 Ex σ0 ~x = 0 = 0 =E 0 0 ~ ext E Physics of Engineered and Nanostructured Materials (SS2011) Sheet5 - 3.6.2011 j0 jx ~ = ρb~j = ρb 0 = 1 −µBj0 E σ0 0 0 ~ x ,E) ~ = tan |Ey | = tan µB ⇒ ∠(E |Ex | c) Here: ~ = (0,0,B) B ~ = (Ex ,Ey ,0) E ~v = (vx ,vy ,0) Equation of motion: e ~ e ~ − ~v ~v˙ = − E − ~v × B m m τ stationary state: vy = 0 (no charges can flow out of the sample), ~v˙ = 0 x-comp of equ of motion: ⇒ vx = − y-comp of equ of motion: ⇒ 0 = − with the cyclotron frequency ωc = eτ Ex m eτ Ey + ωc τ vx m eB m ⇒ Ey = −ωc τ Ex Putting the results from above into the term for the Hall resistivity RH : RH = Uy Ey w Ey Ex ωc τ B = =− = − eτ =− Ix jx wd envx d en m Ex d end The Hall coefficient AH and the Hall resistivity RH are related by: RH (B) = AH ⇒ AH = − B d 1 en (always negative in the Drude model) Additional: Within a solid the mass in the cyclotron frequency equation above is replaced with the effective mass tensor m∗ . Cyclotron resonance is therefore a useful technique to measure effective mass and Fermi surface cross-section in solids. In a sufficiently high magnetic field at low temperature in a relatively pure material 1 τ > kB T ωce > ~ωce where t is the carrier scattering lifetime, kB is Boltzmann’s constant and T is temperature. When these conditions are satisfied, an electron will complete its cyclotron orbit without engaging in a collision, at which point it is said to be in a well-defined Landau level.