**1 Answer**

written 6.0 years ago by |

If a current carrying conductor or semiconductor is placed in a transverse magnetic field, a potential difference is developed across the specimen in a direction perpendicular to both the current and magnetic field. The phenomenon is called HALL EFFECT. As shown consider a rectangular plate of a p-type semiconductor of width ‘w’ and thickness ‘d’ placed along x-axis. When a potential difference is applied along its length ‘a’ current ‘I’ starts flowing through it in x direction.

As the holes are the majority carriers in this case the current is given by

$I=n_h Aev_d$ ……………………………(1)

where $n_h$ = density of holes

A = w × d = cross sectional area of the specimen

$v_d$ = drift velocity of the holes.

The current density is

J= I/A =$n_h ev_d $ ……………………..(2)

The magnetic field is applied transversely to the crystal surface in z direction. Hence the holes experience a magnetic force

$F_m=ev_d B$ …………………………….(3)

In a downward direction. As a result of this the holes are accumulated on the bottom surface of the specimen.

Due to this a corresponding equivalent negative charge is left on the top surface.

The separation of charge set up a transverse electric field across the specimen given by,

$E_H= V_H/d $ …………………………..(4)

Where $V_H$ is called the HALL VOLTAGE and $E_H$ the HALL FIELD.

In equilibrium condition the force due to the magnetic field B and the force due to the electric field $E_H$ acting on the charges are balanced. So the equation (3)

$eE_H=ev_d B$

$E_H=v_d B$ ……………………………….(5)

Using equation (4) in the equation (5)

$V_H=v_d B d $………………………….(6)

From equation (1) and (2), the drift velocity of holes is found as

$v_d= I/(en_h A)= J/(en_h ) $ ……………………..(7)

Hence hall voltage can be written as

$V_H= IBd/(en_h A) = (J_x Bd)/(en_h )$

An important parameter is the hall coefficient defined as the hall field per unit current density per unit magnetic induction.

$R_H= E_H/(J_x B) $