written 8.4 years ago by | • modified 8.4 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 8M
Year: Dec 2013
written 8.4 years ago by | • modified 8.4 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 8M
Year: Dec 2013
written 8.4 years ago by | • modified 8.4 years ago |
$$qE_H = Bqv$$
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}$