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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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qEH=Bqv

Where, q=Magnitude  of  current

v=Drift  velocity.

EH=VHd

But, VHd=Bv

IeVH=B.v.d.(i)

Let n.e be the charge density,

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

v=In.e.w.d(ii)

From (i) and (ii)

VH=B.(In.e.w.d)d

VH=BInew

This is the required expression for Hall voltage.

As, EH=VHd

EH=BJne

EH=BInewd

Dividing both sides by applied electric field (E),

EHE=Bne.(IwdE)

Resistivity,

ρ=RAl

=(VI).Al=ElI.wdl

ρ=wdEI

EHE=Bneρ

ne=EHE.Bρ or ne=EHE.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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