Anonymous

# Magnetic field applied to a semiconductor?

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/m² 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.

Relevance

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)/c², so v(d) = Bc²/E(W)

Then, μE(W) = Bc²/E(W)

μ = Bc²/E²(W)

But E(W) = V(W)/W, so

μ = B(cW/V(W))²

= (0.7) [(3×10^8)(0.004)/(0.0068)]² = 2.18 x 10^16 m²/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/m²