Question

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

The number density of charge carriers is

$n=\frac{JB}{eE}$

Step-by-step solution

Identification of given data

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

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.

F_E=qE

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

$F_B=q\left(V\times B\right)$

Charge density

$J=ne{v}_d$

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

$F_E=qE$ …(i)

And magnetic force as

$F_B=q\left(v_d\times B\right)$

Vis the drift velocity of electrons.

Since V_d and B are perpendicular, we can write

$F_B=q{v}_{d}B$ …(ii)

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

$$\begin{gathered} qE=q{v}_{d}B \\ \therefore v_d=\frac{E}{B} \end{gathered}$$

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

$J=nq{v}_d$

Therefore, number density of charge, n is

$n=\frac{J}{q{v}_d}$

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

$n=\frac{JB}{qE}$