In 1879, Hall found that an e.m.f. is set up transversely or across a current-carrying conductor when a perpendicular magnetic field is applied. This is called the Hall Effect.

To explain the Hall Effect, consider a slab of metal carrying a current. The flow of electrons is in the opposite direction to the conventional current. If the metal is placed in a magnetic field ** B** at right angles to the face AGDC of the slab and directed from CD to AG. Thus electrons accumulate along the side AG of the metal, which will make AG negatively charged and lower its potential with respect to CD. Thus a potential difference or e.m.f. opposes the electron flow. The flow ceases when the e.m.f. reaches a particular value

*V*

_{H}called the Hall voltage, which may be measured by using a high impedance voltmeter.

** Magnitude of Hall Voltage**

Suppose *V*_{H} is the magnitude of the Hall voltage and *d* is the width of the slab. Then the electric field intensity *E* set up across the slab is numerically equal to the potential gradient and hence *E* = *V _{H}/d*. Hence the force on each electron

= *Ee* = *V _{H}e/d.*

This force, which is directed upwards from AG to CD, is equal to the force produced by the magnetic field when the electrons are in equilibrium.

The drift velocity of the electrons is given by

*I = NevA, . . . .*

Where N is the number of the number of electrons per unit volume and A is the area of cross section of the conductor. In this case A = td where *t* is the thickness. Hence, from the second equation above,

v=I/Netd

We now take some typical values for copper to see the order of magnitude of V_{H}. Suppose *B* = 1 T, a field obtained by using a large laboratory electro-magnet. For copper, N is approximately 10^{29} electrons per metre^{3}, and the charge on the electron is 1.6 x 10^{-19} coulomb. Suppose the specimen carries a current of 10 A and that its thickness is about 1 mm or 10^{-3} m. Then

This e.m.f. is very small and would be difficult to measure. The importance of the Hall effect becomes apparent when semiconductors are used, as we now see.

** Hall Effect in Semiconductors**

In semiconductors, the charge carriers which produce a current when they move may be positively or negatively charged. The Hall effect helps us to find the sign of the charge carried. In the diagram above(after the second Paragraph), suppose that electrons were not responsible for carrying the current, and that the current was due to the movement of positive charges in the same direction as the conventional current. The magnetic force on these charges would also be downwards, in the same direction as if the current were carried by electrons. This is because the sign and the direction of movement of the charge carries have both been reversed. Thus **AB** would now become positively charged, and the polarity of the Hall voltage would be reversed. Experimental investigation of the polarity of the Hall voltage hence tells us whether the current is predominantly due to the drift of positive charges or to the drift of negative charges. In this way it was shown that the current in a metal such as copper is due to movement of negative charges, but that in impure semiconductors such as germanium or silicon, the current may be predominantly due to movement of either negative or positive charges.

The magnitude of the Hall voltage *V*_{H} in metals was shown to be very small. In semiconductors it is much larger because the number *N* of charge carried per meter^{3} is much less than in a metal and *V*_{H} = BI/Net. Suppose that *N* is about 10^{25} per meter^{3} in a semiconductor, and B = 1 T (Wb m^{-2}), t = 10^{-3} m, e = 1.6 x 10^{-19} C. then

* V _{H} = *6 x 10

^{-3}V (approx.) = 6 mV.

**It should be noted that the Hall voltage is thus much more measurable in semiconductors than in metals.**

** Use of Hall Effect**

An instrument called a Hall probe may now be used to measure the flux density *B* of a magnetic field. A simple Hall probe is shown in diagram below. Here a wafer of semiconductor has two contacts on opposite sides which are connected to a high impedance voltmeter, V. A current, generally less than one ampere, is passed through the semiconductor and is measured on the ammeter, A. the ‘araldite’ encapsulation prevents the wires from being detached from the wafer. Now, calculating, we have,

Now *Net *is a constant for the given semiconductor, which can be determined previously. Thus from the measurement of *V*_{H} and I, *B* can be found.

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