The Hall Effect

In 1879, E.H. Hall devised an experiment that can be used to identify the sign of the predominant charge carriers in a conducting material. From a historical perspective, this experiment was the first to demonstrate that the charge carriers in most metals are negative.

We investigate the Hall effect by studying the motion of the free electrons along a metallic strip of width l in a constant magnetic field (CNX_UPhysics_28_06_Hall). The electrons are moving from left to right, so the magnetic force they experience pushes them to the bottom edge of the strip. This leaves an excess of positive charge at the top edge of the strip, resulting in an electric field E directed from top to bottom. The charge concentration at both edges builds up until the electric force on the electrons in one direction is balanced by the magnetic force on them in the opposite direction. Equilibrium is reached when:

where e is the magnitude of the electron charge, $v_d$ is the drift speed of the electrons, and E is the magnitude of the electric field created by the separated charge. Solving this for the drift speed results in

A scenario where the electric and magnetic fields are perpendicular to one another is called a crossed-field situation. If these fields produce equal and opposite forces on a charged particle with the velocity that equates the forces, these particles are able to pass through an apparatus, called a velocity selector, undeflected. This velocity is represented in fs-id1171360336437. Any other velocity of a charged particle sent into the same fields would be deflected by the magnetic force or electric force.

Going back to the Hall effect, if the current in the strip is I, then from Current and Resistance, we know that

where n is the number of charge carriers per volume and A is the cross-sectional area of the strip. Combining the equations for $v_d$ and I results in

The field E is related to the potential difference V between the edges of the strip by

The quantity V is called the Hall potential and can be measured with a voltmeter. Finally, combining the equations for I and E gives us

where the upper edge of the strip in CNX_UPhysics_28_06_Hall is positive with respect to the lower edge.

We can also combine fs-id1171360234750 and fs-id1171360485906 to get an expression for the Hall voltage in terms of the magnetic field:

What if the charge carriers are positive, as in CNX_UPhysics_28_06_Hall? For the same current I, the magnitude of V is still given by fs-id1171359011013. However, the upper edge is now negative with respect to the lower edge. Therefore, by simply measuring the sign of V, we can determine the sign of the majority charge carriers in a metal.

Hall potential measurements show that electrons are the dominant charge carriers in most metals. However, Hall potentials indicate that for a few metals, such as tungsten, beryllium, and many semiconductors, the majority of charge carriers are positive. It turns out that conduction by positive charge is caused by the migration of missing electron sites (called holes) on ions. Conduction by holes is studied later in Condensed Matter Physics.

The Hall effect can be used to measure magnetic fields. If a material with a known density of charge carriers n is placed in a magnetic field and V is measured, then the field can be determined from fs-id1171359011013. In research laboratories where the fields of electromagnets used for precise measurements have to be extremely steady, a “Hall probe” is commonly used as part of an electronic circuit that regulates the field.

Velocity Selector

Strategy

Solution

1. The velocity of the unperturbed beam of electrons with crossed fields is calculated by fs-id1171360336437: $v_d=\frac{E}{B}=\frac{6\times {10}^3\text{N}\text{/}\text{C}}{2\times {10}^{\mathrm{-3}}\text{T}}=3\times {10}^6\text{m}\text{/}\text{s.}$
2. The acceleration is calculated from the net force from the magnetic field, equal to mass times acceleration. The magnitude of the acceleration is: $\begin{array}{ccc}ma& =& qvB\\ a& =& \frac{qvB}{m}=\frac{(1.6\times {10}^{\mathrm{-19}}\text{C})(3\times {10}^6\text{m/s})(2\times {10}^{\mathrm{-3}}\text{T})}{9.1\times {10}^{\mathrm{-31}}\text{kg}}=1.1\times {10}^{15}{\text{m/s}}^2.\end{array}$
3. The radius of the path comes from a balance of the circular and magnetic forces, or fs-id1171360336437: $r=\frac{mv}{qB}=\frac{(9.1\times {10}^{\mathrm{-31}}\text{kg})(3\times {10}^6\text{m/s})}{(1.6\times {10}^{\mathrm{-19}}\text{C})(2\times {10}^{\mathrm{-3}}\text{T})}=8.5\times {10}^{\mathrm{-3}}\text{m.}$

Significance

The Hall Potential in a Silver Ribbon

Strategy

Solution

Significance