# Question #e4f6a

Mar 21, 2016

Moving charge carriers in a conductor carrying DC experience a transverse force; when it is placed in a steady magnetic field acting perpendicular to the direction of the current.
Thus there is charge buildup towards one side which creates potential difference.

#### Explanation:

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 $\vec{B}$ is applied as shown.

Electrons, which are the carriers here, experience a transverse force ${\vec{F}}_{m}$ 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 ${\vec{F}}_{e}$ 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 ${\vec{F}}_{M}$ 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 ${\vec{F}}_{H}$ 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 $\vec{B}$ along $z$ axis of a coordinate system. using Lorentz force equation for a charge $q$

$m a t h b f \left\{F\right\} = m a t h b f \left\{{F}_{E}\right\} + m a t h b f \left\{{F}_{M}\right\}$
$= q \cdot \left[\setminus m a t h b f \left\{E\right\} + \left(\setminus m a t h b f \left\{v\right\} \setminus \times \setminus m a t h b f \left\{B\right\}\right)\right]$

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 $E = \frac{V}{d}$, we obtain

${E}_{y} = - {V}_{H} / W$.....(2)

From (1) and (2) we obtain ${V}_{H} = {v}_{x} {B}_{z} W$.......(3)
Recalling that current is given as ${I}_{x} = n \mathrm{dW} \left(- {v}_{x}\right) \left(- e\right)$
where $n$ is charge carrier density, $d \times W$ 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
${V}_{H} = {I}_{x} / \left(n \mathrm{dW} e\right) {B}_{z} W = \frac{{I}_{x} {B}_{z}}{n \mathrm{de}}$