Question

What is Hall Effect? Deduce an expression for Hall coefficient.​

Answer 1

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.

1

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

I=nhAevd ……………………………(1)

where nh = density of holes

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

vd = drift velocity of the holes.

The current density is

J= I/A =nhevd ……………………..(2)

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

Fm=evdB …………………………….(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,

EH=VH/d …………………………..(4)

Where VH is called the HALL VOLTAGE and EH the HALL FIELD.

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

eEH=evdB

EH=vdB ……………………………….(5)

Using equation (4) in the equation (5)

VH=vdBd………………………….(6)

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

vd=I/(enhA)=J/(enh) ……………………..(7)

Hence hall voltage can be written as

VH=IBd/(enhA)=(JxBd)/(enh)

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

RH=EH/(JxB)

Answer 2

Hall Effect: It is the ratio of an induced electric field to the product of the current density and the applied magnetic field.

This involves the production of a voltage difference across a conductor that is transverse to an electric current and magnetic field which is perpendicular to the field.

Here holes being the major charge carriers, we get current as,

⇒ I = $n_hAev_d$ → (equation 1)

Where $n_h$ is the density of holes, A = w×d = cross-sectional area, and $v_d$ is the drift velocity of the holes.

The current density is given as,

⇒J = I/A = $n_hev_d$ → (equation 2) ( obtained from equation 1)

Here holes experience a magnetic field as it is applied transversely to the plane,

∴ $F_m=ev_dB$ → (equation 3)

The magnetic field is in the downward direction therefore the holes are accumulated in the below surface.

There is an occurrence of electric filed due to separation of charge,

∴$E_H$ = $V_H/d$ → (equation 4)

At equilibrium both electric field and magnetic field are balanced,

∴ equation 3 becomes,

$eE_H = ev_dB$

$E_H=v_dB$ → (equation 5)

Using equation 4 and in equation 5 we get,

$V_H=v_dBd$  → (equation 6)

We can find the drift velocity from Equations 1 and 2.

$v_d=I/(en_hA)=J/(en_h)$ → (equation 7)

Hence Hall Voltage is,

$V_H = IBD/(en_hA)=(J_xBD)/(en_h)$

We know that the Hall coefficient is defined as the hall field per unit current density per unit magnetic induction.

∴$R_H = E_H/(J_xB)$