Electronic Properties of Materials: Conduction - Problems and Solutions

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Consider a rectangular sample, a metal or an n-type semiconductor, with a length L, width W, and thickness D. A current I is passed along L, perpendicular to the cross-sectional area WD. The face W X L is exposed to a magnetic field intensity B. A voltmeter is connected across the width to read the Hall voltage VH.

  1. Show that the Hall voltage recorded by the voltmeter us VH = IB/Den
  2. Consider a 1-micron-thick strip of gold layer on an insulating substrate that is a candidate for a Hall probe sensor. If the current through the film is maintained at constant 100 mA. What is the magnetic field that can be recorded per micro-volt of Hall voltage?



  1. Consider a slab of length L, width W, and thickness D as shown above. Let us take a small element "dX" in the length; its volume is given by dV = WDdX
    If n = number of charge carriers per unit volume, then the total number of charge carriers in this volumetric element of thickness dX will be nWDdX
    Therefore, total charge (dQ) = enWDdX
    Thus, current (I) = dQ/dt = enWD(dX/dt)
    Or, I = enWDv (where v is the charge velocity)
    Or, ne = I/WDv
    Magnetic force (Fm) = q(v x B)
    Or, Fm = nevB = (I/WDv) x (vB) = IB/WD

    Electrostatic force (Fe) = qE = neE
    A potential develops across the width W, and is given by VH = EW. VH is called as the Hall potential.
    It gives the value of the electric field E = VH/W
    Therefore, Fe = neVH/W

    In equilibrium, Fe = Fm
    That is, neVH/W = IB/WD
    Or, VH = IB/neD
    This equation shows that the Hall potential is independent of the length and the width of the conductor slab.

  2. The atomic weight of gold is 196.97 grams.
    i.e., 6.023 x 1023 atoms weigh 196.97 grams
    or, 1 gram gold has 6.023 x 1023/196.97 atoms

    The density of gold is 19.32 g/cm3.
    i.e., 1 cm3 of gold weighs 19.32 grams.
    Therefore, 1 cm3 has [6.023 x 1023/196.97] x 19.32 atoms
    i.e., n = 5.9077 x 1022 atoms/cm3 or 5.9077 x 1028 atoms/m3

    Gold is monovalent (valency = 1), therefore, total number of conduction electrons = total number of atoms
    Thus, total number of conduction electrons per m3 = 5.9077 x 1028.
    Therefore, total charge = ne = 5.9077 x 1028 x 1.6 x 10-19

    VH = IB/neD
    Or, B/VH = neD/I = 10-6 x 5.907728 x 1.6 x 10-19 / 100 x 10-3 (Given thickness = 1 micron and current = 100 mA)
    B/VH = 94523.2 Tesla/Volt
    Or, B/VH = 94523.2 x 10-6 / 10-6 Tesla/Volt
    Or, B/VH= 0.0945232 Tesla/micro Volt