Advertisement Remove all ads

State the Hall Effect. Derive the Expression for Hall Voltage and Hall Coefficient with Neat Diagram. - Applied Physics 1

Answer in Brief

State the Hall effect. Derive the expression for Hall voltage and Hall coefficient with neat diagram.

State the Hall effect. Derive the expression for Hall coefficient with neat diagram.

Advertisement Remove all ads

Solution

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 = nh Aevd  ……………………………(1)

where n= density of holes

A  = w×d =  crosss ectional area of the specimen

 Vd = 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

   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,

`E_ = (V_H)/ 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) 

`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. A) = ( J_xBd)/( en.)` as

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_xB)`


Concept: Hall Effect
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×