A rectangular plate of semiconducting material has dimensions 10mm x 4mm x 1mm, a current of 1.5mA flows along its length and the associated potential drop is 78mV. When a magnetic field of strength 0.7 Wb/m? is applied normally to the major surface of the sample a potential difference of 6.8mV appears across the width of the sample. Determine the character, concentration and mobility of the current carriers.
Given that the semiconductor is L x W with a current, I, flowing along L creating a potential drop, V(L) across L. Now a magnetic field B is applied normal to the major surface and creating a potential difference of V(W) across W. Hence, the semiconductor has a resistance, R given by: R V(L)/I(L) (.078)/(0.0015) 52 Ohms Now the mobility of the current carrier is given by the drift velocity: v(d) μE(W) where μ is the mobility and E(W) is the electric field across W Also, B v(d) E(W)/c?, so v(d) Bc?/E(W) Then, μE(W) Bc?/E(W) μ Bc?/E?(W) But E(W) V(W)/W, so μ B(cW/V(W))? (0.7) [(3×10^8)(0.004)/(0.0068)]? 2.18 x 10^16 m?/V·sec To determine the charge concentration Q/A, where A LW We first find the current along W I(W) V(W)/R (0.0068)/50 0.136 mA So we know that the Lorentz force due to the magnetic field B is given by F I(L)LB, and it is also BQv(d) So I(L)L Qv(d) but v(d) μE(W) μV(W)/W So I(L)L QμV(W)/W Now rewrite the above equation as: Q/A Q/LW I(L)/μV(W) (0.136)/(2.18x10^16)(0.0068) 9.17 x 10^-16 C/m?
4 wheels work, but on the different levels, it depends on the surface. One can stop at all because it is on the ice and so on. But only front wheels work with the ABS
