As stated above, Hall effect is observed even when there is steady state. Clearly it is a phenomenon different from induction which results from varying current or varying magnetic field.
Lets understand the production of Hall Voltage in a flat conducting plate as shown in the picture below.
Let a current I flow through a conducting plate of thickness d. A perpendicular magnetic field vecB is applied as shown.
Electrons, which are the carriers here, experience a transverse force vecF_m due to the magnetic field and are pushed to one edge of the plate. This leaves the other edge deficient in electrons or positively charged and an electrical field vecF_e is created in the plate as shown. Thus potential difference V_H, called Hall Voltage is created between the two edges. Which can be measured with the help of a sensitive voltmeter.
Direction of magnetic force vecF_M can be ascertained with the help of right hand rule. One needs to remember that electrons are negatively charged and therefore, their velocity is in a direction opposite to that of the conventional current I. This also fixes the direction of electric field vecF_H produced due to Hall effect.
Hall effect is noticed in semiconductors as well. For n type semi conductors carriers being negatively charged electrons treatment is similar. In p type semi conductors carriers are considered holes which carry positive charge and are dealt with appropriately.
Calculations of Hall voltage:
Let W is the width of the conducting plate, current I be along x axis, vec F_e along y axis and vecB along z axis of a coordinate system. using Lorentz force equation for a charge q
mathbf{F} = mathbf{F_E}+mathbf{F_M}
=qcdot[\mathbf{E} + (\mathbf{v} \times \mathbf{B})]
Under steady state conditions no charges are moving in the y-axis direction. As the magnetic force on each electron in the y-axis direction is cancelled by an y-axis electrical force. The expression reduces to
0 = E_y - v_x B_z , ......(1)
where -v_x is the drift velocity of electrons constituting current I. Also noting that for a contant electric field E=V/d, we obtain
E_y=-V_H/W.....(2)
From (1) and (2) we obtain V_H=v_xB_zW.......(3)
Recalling that current is given as I_x=ndW(-v_x)(-e)
where n is charge carrier density, d xx W is area of cross-section for the current flow, and -e is the charge of each electron. Substituting value of v_x in (3) we obtain
V_H=I_x/(ndWe)B_z W=(I_xB_z)/(nde)
