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In a Hall-effect experiment, express the number density of charge carriers in terms of the Hall-effect electric field magnitude E, the current density magnitude J, and the magnetic field magnitude B.

Short Answer

Expert verified

The number density of charge carriers is

n=JBeE

Step by step solution

01

Identification of given data

  1. Electric field magnitude is E .
  2. Magnetic field magnitude is B.
  3. Charge density magnitude J.
02

Significance of magnetic force

The influence of a magnetic field produced by one charge on another is what is known as the magnetic force between two moving charges

We can write the electric force and magnetic force equations and equate them since these forces balance each other. Further, using the formula for current density, we can solve it for number density for charge carriers.

Formula:

Electric force on charge q in electric field E.

FE=qE

Magnetic force on charge q moving with velocity V in field B

FB=qV×B

Charge density

J=nevd

03

Step 3: Express the number density of charge carriers in terms of E, j, and B

In Hall Effect, electric force and magnetic force balance each other.

We can write the electric force as

FE=qE …(¾±)

And magnetic force as

FB=qvd×B

Vis the drift velocity of electrons.

Since Vd and B are perpendicular, we can write

FB=qvdB …(¾±¾±)

Equating equation (i) and (ii), we get

qE=qvdB∴vd=EB

…(iii)Now, current density can be written as

J=nqvd

Therefore, number density of charge, n is

n=Jqvd

Substituting the value of vd from equation (iii) in the above equation, we get

n=JBqE

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