The Hall Effect

A current in a conductor means moving charges and we know that a magnetic force exerts force on moving charges. The Hall effect demonstrates the effect of magnetic force on these moving charges within the conductor.

To illustrate this effect we consider a flat conductor placed in a uniform magnetic field. The magnetic field is perpendicular to the plane of the conductor and directed into the screen (your computer's screen or phone's screen).

In Figure 1 the moving charge is positive. The magnetic field exerts upward force on the charge and the upper region of the conductor is positively charged and the lower region negatively charged. In Figure 2, the moving charge is negative and also experiences the upward magnetic force.

Figure 1 Hall effect; moving charge is positive.
Figure 2 Hall effect; moving charge in negative

Note that the magnetic force is upward for both positive and negative charges, that is the magnetic force is in the same direction for a current carrying conductor regardless of the positive or negative charges. Do not confuse with the direction of magnetic force here. If you apply the right hand rule to find the direction, the direction for negative charge is opposite to the direction the thumb points.

Let's consider the case in Figure 1 where a positive charge moves with drift velocity $\vec v_d$. The electric field is established due to the separation of charges in the upper and lower region of the conductor and it exerts electric force in opposite direction. The separation of charges continues until the upward magnetic force $\vec F_B$ is balanced by the downward electric force $\vec F_E$.

Therefore, a potential difference is created between the upper region and lower region of the conductor, that is between points $a$ and $b$ in Figure 1 and Figure 2 called Hall emf. We can prove by detecting the polarity of the conductor that the moving charges are indeed electrons in metals.

We can also prove that the conduction in semiconductors takes place by the conduction of positively charged holes, that is the Hall emf in semiconductors is opposite to the Hall emf in metals. A hole is created if an electron leaves its place and the empty place acts as a positive charge.

The magnitude of electric force on the charge $q$ is $qE$ and the magnitude of magnetic force is $qv_dB$. At the balanced condition we have $qE + qv_dB = 0$ and therefore,

\[E = -v_d\,B\]

If $d$ is the width of the conductor, the potential difference $V_{ab}$ of point $a$ with respect to point $b$ (Hall voltage or emf) is

\[V_{ab} = -v_dB\,d\]

We know that the current density $J$ is $J = nqv_d$ where $n$ is the concentration of charge (charge per unit volume) and $v_d = J/nq$, so the hall voltage is

\[V_{ab} = V_H = -\frac{JBd}{nq} \]

The negative sign is valid for both positive and negative charges. If $q$ is negative, the potential difference is positive and vice versa.

The quantity $1/nq$ is called Hall coefficient. We can determine the value of Hall coefficient by measuring $J$, $B$ and $V_H$. That allows you to directly measure the concentration $n$ of the electrons in metals and semiconductors. We already know the magnitude of charge $q$ which is the magnitude of charge of an electron for both metals and semiconductors.

The measurement of concentration $n$ with this approach is nearly valid for alkali metals which give one electron for conduction such as Li, Na, K and for some transition metals Cu and Ag. And the measurement of $n$ for semiconductors is more complex. The Hall effect is classical approach and quantum physics is required to be fully in agreement with the experimental results.

Another interesting fact is you can directly measure the drift velocity $v_d$ of electrons. Move the conductor in the opposite direction of current until Hall emf is zero and the speed of motion of the conductor gives the magnitude of the drift velocity. If you move the conductor, the electrons are at rest for the magnetic field when the Hall emf is zero. The drift velocity appears to be very slow, that is of the order of $10^{-4} \text{m/s}$.

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