## What is Hall Effect?

### Derive an expression for Hall coefficient, Hall potential, Hall Voltage, and Hall angle. What is the **importance of the Hall effect?**

Answer:

## Hall Effect

The Hall Effect discovered by Edwin Hall in 1879. He states that “when we flow a current along the length of the conducting ribbon (conductor or semiconductor) and placed it into the transverse magnetic field, then an electric potential developed across the ribbon. The direction of electric potential is perpendicular to applied current ** I** and magnetic field

**. The electric potential is known as Hall potential and generated field is known as Hall Field and we know this effect as “**

*B**”.*

**Hall Effect**If we take a conducting ribbon flowing current in +ve X-direction and the magnetic field is applied in the +ve Z-direction. A force is experienced on charge carrier in the downward direction of conducting ribbon means –ve Y- direction. Through this force, the +ve charges arranged at the upper edges and –ve charges arranged at the bottom edges of the ribbon. The separation of charge carriers developed the electric field inside the ribbon in the direction of Y.

If we consider a ribbon has a width W and thickness is T. There are two forces acts on electron due to electric field and magnetic field

i.e.

Electric field force on electron = -eE_{H } (1)

Magnetic field force on electron = -eV_{x}B_{Z }(2)

_{ }In equilibrium condition, -eE_{H }= -eV_{x}B_{Z}

E_{H }= V_{x}B_{Z} (3)

Now we define current density (Current density is a measure of the density of an electric current.) toward X i.e.

J_{x }= -neV_{x } ( n = no. of charge carrier / unit Volume)

V_{x} = – J_{x }/ ne (4)

Fron equation (4) and (3)

E_{H }= -B_{Z} J_{x }/ ne_{ = }-(1/ne) B_{Z} J_{x} (5)

R_{H} = -(1/ne) ** **(6)

#### R_{H} is known as *Hall Coefficient*.

*Hall Coefficient*.

E_{H }= R_{H} B_{Z} J_{x}

R_{H }= E_{H }/ B_{Z} J_{x }(7)

If we take length **L** and breadth **b** of the conducting ribbon and hall field E_{H} and developed potential is V_{H}

Then E_{H }= V_{H} / L (8)

R_{H }= V_{H} / L B_{Z} J_{x}

V_{H} = R_{H} L B_{Z} J_{x }(9)

Again current density is equal to the current per unit area

i.e J_{x } = I_{x}/ LB

V_{H} = R_{H} L B_{Z} I_{x}/ LB

**V _{H} = R_{H} B_{Z} I_{x}/ B (10)**

**V**_{H } is known as *Hall Potential.*

_{H }is known as

*Hall Potential.*

**Calculation of Hall angle and Mobility of charge carrier**:

The charge carrier mobility is equal to the drift velocity per unit electric field

i.e. µ = V_{x}/ E

Since for equation (3) E_{H }= V_{x}B_{Z }

then E_{H }= µEB_{Z}

E_{H }= µEB_{Z}

Since E_{H }= R_{H} B_{Z} J_{x}

Compare both equation µEB_{Z } = R_{H} B_{Z} J_{x}

µ= R_{H} J_{x}/ E (11)

**Mobility µ= ****s****R _{H}** (s = J

_{x}/ E current density per unit electric field)

Since E_{H }= µEB_{Z}

µ= E_{H} / EB_{Z}

q_{H} = E_{H} / E is known as *Hall angle*.

** **

**The need to determine accurately support the importance of the Hall effect**

- carrier density.
- electrical resistivity.
- and the mobility of carriers in
*semiconductors.* - Hall Effect proved that electrons are the majority carriers in all the metals and n-type semiconductors.
- In p-type semiconductors, holes are the majority carriers.

**Problem 1: Find the Hall coefficient for 5 x 10**^{28} atom / m^{3 }in the copper block.

^{28}atom / m

^{3 }in the copper block.

**Solution:** R_{H} =-1/ne

R_{H} = -1/5 x 10^{28 }x 1.6 x 10^{-19}

R_{H} = -0.125 x 10^{-9} m^{3}/C

**Problem 2: Calculate mobility and charge carrier density when the resistivity of doped Si sample is 9 x 10**^{-3} Ω–m and the Hall coefficient is 3.6 x 10^{-4} m^{3}/C.

^{-3}Ω–m and the Hall coefficient is 3.6 x 10

^{-4}m

^{3}/C.

**Solution:** Given that r =9 x 10^{-3} Ω–m; R_{H} = 3.6 x 10^{-4} m^{3}/C

So conductivity s = 1 / r = 1/ 9 x 10^{-3} /Ω–m

s = 111.1 / Ω–m

r = ne;

n = r/e = 1/ R_{H}e

n = 1/ 1.6 x 10^{-4} x 1.6 x 10^{-19}

n = 1.7363 x 10^{22 }/ m^{3}

Mobility µ = R_{H }s

µ = 111.1 x 3.6 x 10^{-4}

µ = 0.04 m^{2}/V-s

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