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Explain the hall effect in metal ? Derive the formulae to determine the density and mobility of the electrons.
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Hall Effect –

If a metal or semiconductor carrying current I is placed in transverse magnetic field B, an electric field E is induced in the direction perpendicular to both I & B.

This phenomenon is known as Hall Effect & the electric field or voltage induced is called voltage $(V_H)$.

Hall Effect is very important in semiconductor because it can be used to determine –

1) The nature (P-type or N-type) of semiconductor

2) To measure charge carrier concentration.

3) To measure the mobility of charge carrier if the conductivity of material is known in magnetic field meter also.

Derive –

1) In equilibrium, the electric field intensity $E_H$ due to Hall Effect exerts a force on charge carrier which just balances the magnetic force.

$\therefore qE_H =BqV $ or $E_H=BV$

Where $q=$ magnitude of charge on carrier

$V=$ drift speed

2)  We know that,

$E_{H}=\dfrac{V_{H}}{d}$

3)  If $J=$ current density, $n\cdot e =$ charge density & $w=$ specimen width, then

$J=n\cdot e\cdot v=\dfrac {I}{wd}$

4) $\therefore V_{H}=(E_{H})\cdot d=(BV)\cdot d=\dfrac {BJd}{n\cdot e}=\dfrac {BI}{n\cdot e\cdot w} $

5) $\because E_{H}=\dfrac {V_{H}}{d}=BV=\dfrac{BJ}{n\cdot e}$

6) Now $J=n\cdot e\cdot V =\dfrac {I}{A}$ also, $J=\sigma E$

$\therefore \sigma = \dfrac {J}{E}= \dfrac {I}{A}\cdot \dfrac {1}{E}$.....(a)

7) $\therefore E_{H}=\dfrac {BJ}{n\cdot e}=\dfrac {B}{n\cdot e}\left [\dfrac {I}{A} \right ]=\dfrac {B}{n\cdot e}\left( \dfrac {I}{A} \cdot \dfrac {E}{E} \right )$.....(multiplying and dividing by $E$) $\therefore \dfrac {E_{H}}{E}=\dfrac{B}{ne}\times \left (\dfrac {I}{A}\times \dfrac {1}{E} \right )= \dfrac {B}{ne}\cdot \sigma$.....from (a) 8)  We know that, $\rho =\dfrac {1}{\sigma}$ $\sigma = \dfrac {1}{\rho}$ $\therefore \dfrac {E_{H}}{E}=\dfrac {B}{ne}\cdot \sigma$ thus, $\dfrac {E_{H}}{E}=\dfrac {B}{ne \rho}$ $\therefore n=\left(\dfrac {E}{E_{H}} \right )\times \left (\dfrac {B}{e\rho} \right )$ 9) We know that, $\mu=\dfrac {V_{d}}{E}$ but $E_{H}=BV_{d}$ $\therefore V_{d}= \dfrac {E_{H}}{B}$ $\therefore \mu =\dfrac {E_{H}}{E\times B}$

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