written 8.7 years ago by | • modified 8.7 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 8M
Year: Dec 2013
written 8.7 years ago by | • modified 8.7 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 8M
Year: Dec 2013
written 8.7 years ago by | • modified 8.7 years ago |
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}