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Explain Hall Effect in metal. Derive the formulae to determine the density and mobility of electrons.

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 8M

Year: Dec 2013

1 Answer
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$$qE_H = Bqv$$

Where, $q = Magnitude \ \ of \ \ current$

$v = Drift \ \ velocity.$

$E_H = \frac{V_H}{d}$

But, $\frac{V_H}{d} = Bv$

$IeV_H = B.v.d….(i)$

Let n.e be the charge density,

Now, $ I = (n.e)(w.d).v$

$ v = \frac{I}{n.e.w.d} …(ii)$

From (i) and (ii)

$V_H = B. (\frac{I}{n.e.w.d})d$

$V_H = \frac{BI}{new}$

This is the required expression for Hall voltage.

As, $E_H = \frac{V_H}{d}$

$E_H = \frac{BJ}{ne}$

$E_H = \frac{BI}{newd}$

Dividing both sides by applied electric field (E),

$\frac{E_H}{E} = \frac{B}{ne}.(\frac{I}{wdE})$

Resistivity,

$ ρ = \frac{RA}{l}$

$ = (\frac{V}{I}).\frac{A}{l} = \frac{El}{I}.\frac{wd}{l}$

$ ρ = \frac{wdE}{I}$

$\frac{E_H}{E} = \frac{B}{ne ρ}$

$ne = \frac{E_H}{E}.\frac{B}{ρ}$ or $ ne = \frac{E_H}{E}.Bσ$ [ σ is conductivity]

This is the required expression for charge density.

Conductivity ( σ = 1/ρ ) is related to mobility(µ) as follows,

$σ = µne$

$\frac{σ}{ne} = µ --------(A)$

Hall co-efficient is defined as,

$R_{H=1/ne}$

$R_{H=µ/σ}$ (from A)

$µ = σ R_H -------(B)$

Also,

$ R_H = \frac{V_H w}{BI}$

$ µ = \frac{σ V_H w}{I}$

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