A bar of n-type Ge of size 0.010m × 0.001m × 0.001m is mounted in a magnetic field of 0.2 T. The electron density in the bar is 7 × $10^{21}/m^3$. If 1 mV is applied across the long ends of the bar,Assume $μ_e = 0.39 m^2/Vs$. -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: May 2013

Given:-

$l = 0.01m, w = 10^{-3}m , d = 10^{-3}m$

Magnetic field, B=0.2T

Electron density, $n = 7 × 10^{-21}/m^3$

Potential difference, $V_A = 1mV = 10^{-3}V$

$µ_e = 0.39m^2/Vs$

To find:-

(i) Current (I)

(ii) Hall voltage$(V_H)$

Solution:-

Conductivity is given as,

$σ = µne$

Resistance of conductor,

$R = \frac{l}{σ A} = \frac{l}{σ.wd} = \frac{l}{µ.n.e.w.d}$

By Ohm’s law,

$I = \frac{V_A}{R} = \frac{V_A µ.n.e.w.d}{l}$

$ = \frac{10^{-3} × (0.39) × (1.6 × 10^{-19}) × 7 × 10^{21} × 10^{-3} × 10^{-3}}{0.01}$

$I = 4.368 × 10^{-5}A$

Or $ I = 43.68µA$ Ans

Now, $V_H = \frac{BI}{new} = \frac{0.2 × 4.368 × 10^{-5}A}{7 × 10^{21} × 1.6 × 10^{-19} × 10^{-3}} = 7.8 × 10^{-3}V$

$V_H = 7.8mV$