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Determine the current through the bar and the voltage between the Hall electrodes placed across the short dimensions of the bar.

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 × 1021/m3. If 1 mV is applied across the long ends of the bar,Assume μe=0.39m2/Vs. -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: May 2013

1 Answer
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Given:-

l=0.01m,w=103m,d=103m

Magnetic field, B=0.2T

Electron density, n=7×1021/m3

Potential difference, VA=1mV=103V

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

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