Crossed Fields: The Hall Effect — Question 2

Question 2

A measured Hall voltage determines carrier density in a one-carrier model

A strip of thickness .500 mm carries I=20.0 mA along +x in B=.500 T along +z. The upper-minus-lower Hall voltage is +.250 mV. Assume singly charged carriers with magnitude $e=1.60\times10^{-19}$ C.

1. Find the carrier sign and density.

2. For strip width 4.00 mm, find the signed drift velocity and transverse electric field.

Question 2 - Solution

A measured Hall voltage determines carrier density in a one-carrier model

(a) Method and calculation

$$V_H=-IB/(nqt)>0\Rightarrow q=-e.$$$$n=\frac{IB}{et|V_H|}=\frac{(.0200)(.500)}{(1.60\times10^{-19})(.000500)(.000250)}=5.00\times10^{23}\,\mathrm{m^{-3}}.$$

(b) Interpretation and further calculation

$$v_d=\frac I{nqwt}=-.125\,\mathrm{m/s}.$$$$E_y=v_dB=-.0625\,\mathrm{V/m}=-V_H/w.$$

Physical check

The sign is fixed by the stated current, field and voltage-contact directions; the density is positive.