Answer to Question #160878 in Electric Circuits for charan

Question

Rh=1/resistivity

Why hall coefficient is equal to 1/restivity why this relation come from

what is the relation between resistivity and hall coefficient may I get the relation with proof

Answer 1

In wires, electrons instead of holes are flowing, so $vx → −vx \space and \space q → −q. \space \space Also \space \space Ey = − VH / w$

${\displaystyle V_{\mathrm {H} }=v_{x}B_{z}w}$

The conventional "hole" current is in the negative direction of the electron current and the negative of the electrical charge which gives $Ix = ntw(−vx)(−e)$  where n is the charge carrier density, tw is the cross-sectional area, and −e is the charge of each electron. Solving for ${\displaystyle w}$ and plugging into the above gives the Hall voltage:

${\displaystyle V_{\mathrm {H} }={\frac {I_{x}B_{z}}{nte}}}$

If the charge build-up had been positive (as it appears in some semiconductors), then the V_H assigned in the image would have been negative (the positive charge would have built up on the left side).

The Hall coefficient is defined as

${\displaystyle R_{\mathrm {H} }={\frac {E_{y}}{j_{x}B_{z}}}} \space or \space {\displaystyle E=-R_{H}(J_{c}\times B)}{\displaystyle E=-R_{H}(J_{c}\times B)}$

where j is the current density of the carrier electrons, and E_y is the induced electric field. In SI units, this becomes

${\displaystyle R_{\mathrm {H} }={\frac {E_{y}}{j_{x}B}}={\frac {V_{\mathrm {H} }t}{IB}}={\frac {1}{ne}}.}$

(The units of R_H are usually expressed as m³/C, or Ω·cm/G, or other variants.) As a result, the Hall effect is very useful as a means to measure either the carrier density or the magnetic field.