The Hall Effect and the Hall Probe

A2 · 14 min

When a current flows through a thin slice of material in a magnetic field, a small voltage appears across the slice, at right angles to the current. This is the Hall effect. It comes straight from the force $BQv$ on the moving charge carriers, and it is the working principle of the Hall probe, the standard instrument for measuring magnetic flux density. The syllabus requires you to explain where the Hall voltage comes from, to derive $V_H = BI/ntq$ , and to use a Hall probe; the derivation is a common 3 to 4 mark Paper 4 question.

Setting the scene

Take a thin rectangular slice of a conductor or semiconductor. Give it three dimensions:

length: the direction the current $I$ flows,
width $d$ : across the face of the slice,
thickness $t$ : the small dimension, parallel to the magnetic field.

A uniform magnetic field $B$ acts at right angles to the large face of the slice (so it passes through the thickness). The current consists of charge carriers, each of charge $q$ , drifting along the slice with drift speed $v$ . There are $n$ carriers per unit volume.

A Hall slice seen face-on. Conventional current flows to the right and B is into the page. The electrons (moving left) are pushed to the top edge, leaving the bottom edge positive. The Hall voltage V_H is measured across the width d.

Where the Hall voltage comes from

Follow the charge carriers step by step. Take a metal, where the carriers are electrons, with conventional current to the right and $B$ into the page.

The electrons drift to the left (opposite to the conventional current).
Each electron feels a magnetic force $Bqv$ , perpendicular to its velocity and to the field. Fleming's left-hand rule (second finger along the conventional current, to the right) gives a force towards the top edge.
Electrons pile up on the top edge, which becomes negative. The bottom edge is left with a shortage of electrons and becomes positive.
These separated charges set up an electric field across the width of the slice, from the positive edge to the negative edge. This field pushes each electron back towards the bottom with a force $qE$ .
As more charge builds up, the electric force grows. Very quickly it becomes equal to the magnetic force. Then the electrons feel no resultant sideways force and drift straight along the slice again.
In this steady state, there is a constant p.d. across the width: the Hall voltage $V_H$ .

Definition

The Hall voltage $V_H$ is the potential difference produced across a current-carrying conductor (or semiconductor), at right angles to both the current and an applied magnetic field, when the charge carriers are deflected to one side by the magnetic force.

Deriving the Hall voltage equation

Step 1: balance the forces. In the steady state the electric force on a carrier equals the magnetic force on it. The electric field across width $d$ is $E = V_H/d$ , so

qE = Bqv \quad\Rightarrow\quad \frac{qV_H}{d} = Bqv \quad\Rightarrow\quad V_H = Bvd

Step 2: eliminate the drift speed. From AS, $I = nAvq$ . The cross-sectional area through which the current flows is width times thickness, $A = dt$ , so

I = n\,d\,t\,v\,q \quad\Rightarrow\quad v = \frac{I}{ndtq}

Step 3: substitute.

V_H = B\left(\frac{I}{ndtq}\right)d = \frac{BI}{ntq}

Key result

V_H = \frac{BI}{ntq}

$V_H$ Hall voltage (V), $B$ flux density (T), $I$ current (A), $n$ number density of charge carriers ( $\text{m}^{-3}$ ), $t$ thickness of the slice in the direction of the field (m), $q$ charge of each carrier (C).

The width $d$ cancels. The Hall voltage depends on the thickness $t$ , the dimension parallel to the field, not on the width across which the voltage is measured.

What makes a large Hall voltage

From $V_H = BI/ntq$ , for a given field the Hall voltage is larger when:

the current $I$ is larger,
the slice is thinner (smaller $t$ ),
the number density of charge carriers $n$ is smaller.

The last point is why Hall probes use semiconductors, not metals. A metal has $n \approx 10^{28}$ to $10^{29}\ \text{m}^{-3}$ ; a semiconductor has perhaps $10^{20}$ to $10^{23}\ \text{m}^{-3}$ . For the same current, fewer carriers must drift much faster ( $I = nAvq$ ), so the magnetic force on each is larger and so is the Hall voltage. In a metal $V_H$ is typically microvolts, too small to measure easily; in a semiconductor it is millivolts.

Tip

The sign of the Hall voltage reveals the sign of the charge carriers. Positive carriers drifting the same way as the conventional current are pushed to the same edge as electrons drifting the opposite way, but they make that edge positive. Measuring which edge is positive shows whether the carriers are electrons or positive "holes", as they are in p-type semiconductors. This is extension material, but it explains why some materials have Hall voltages of the "wrong" sign.

The Hall probe

A Hall probe is a thin slice of semiconductor at the tip of a handle, with a constant current supplied to it and connections across its width to measure $V_H$ .

For a given probe, $I$ , $n$ , $t$ and $q$ are all constant, so

V_H \propto B

The Hall voltage is directly proportional to the flux density. A probe is calibrated by placing it in a known field (for example the centre of a long solenoid, or a field measured by a current balance) and finding the constant $V_H/B$ . After that, any reading of $V_H$ gives $B$ directly. Many probes have a meter that reads in tesla.

Orientation matters

Only the component of $B$ perpendicular to the face of the slice deflects the carriers across the width. If the normal to the face is at angle $\alpha$ to the field, the reading is proportional to $B\cos\alpha$ .

Measuring a flux density with a Hall probe

Calibrate the probe in a known field, or use a probe with a calibrated scale. Set the reading to zero away from all magnets (or note the reading with no field present, to allow for the Earth's field and zero error).
Place the probe at the point of interest with its flat face perpendicular to the field.
Rotate the probe slowly until the reading is a maximum. At the maximum the face is perpendicular to the field, so this gives the full value of $B$ .
Read $V_H$ and convert to $B$ using the calibration constant: $B = V_H/k$ .
To find the direction of the field, note the orientation at which the reading is maximum (the field is along the normal to the face) and the sign of the reading.

Using a Hall probe to investigate a magnetic field

Typical investigations: how $B$ varies with distance $r$ from a long straight wire; how $B$ at the centre of a solenoid depends on the current; how $B$ varies along the axis of a coil.

Apparatus: calibrated Hall probe and meter (or probe, constant-current supply and millivoltmeter); the current-carrying wire or coil with a d.c. supply, ammeter and variable resistor; metre rule or a clamp stand with a scale to position the probe; set squares to align the probe.

Method (wire):

Set a constant current, for example $5\ \text{A}$ , in a long vertical straight wire. Measure the current with an ammeter.
Clamp the probe with its face perpendicular to the circular field lines (its face contains the wire's direction and the radial line).
Measure the distance $r$ from the centre of the wire to the sensor in the probe. Record $V_H$ or $B$ .
Repeat for at least six distances, and with the current reversed; average to cancel the Earth's field.

Analysis: theory predicts $B \propto 1/r$ for a long straight wire. Plot $B$ against $1/r$ : a straight line through the origin confirms it.

Control variables: the current; the orientation of the probe; keep other magnets and steel objects away.

Sources of error and improvements:

The Earth's field (about $5 \times 10^{-5}\ \text{T}$ ) adds to readings: zero the probe with no current, or reverse the current and average.
The sensor position inside the probe is uncertain: measure distances to a marked sensor position, or plot a graph so a constant offset shows as an intercept.
The probe face may not be exactly perpendicular: rotate to the maximum reading at each position.
The wire heats and the current falls: monitor the ammeter and adjust.

Worked examples

Hall voltage in copper

A copper strip $0.10\ \text{mm}$ thick carries a current of $5.0\ \text{A}$ at right angles to a magnetic field of $0.50\ \text{T}$ that is perpendicular to its face. Copper has $8.5 \times 10^{28}$ free electrons per cubic metre. Calculate the Hall voltage and comment on its size.

Solution

V_H = \frac{BI}{ntq} = \frac{0.50 \times 5.0}{8.5 \times 10^{28} \times 0.10 \times 10^{-3} \times 1.60 \times 10^{-19}} = 1.8 \times 10^{-6}\ \text{V}

About $2\ \mu\text{V}$ , even with a large current and a strong field. This is far too small for a practical probe, which is why semiconductors are used.

Carrier density in a semiconductor

A slice of semiconductor of thickness $0.50\ \text{mm}$ carries a current of $20\ \text{mA}$ . In a field of $0.060\ \text{T}$ perpendicular to its face, the Hall voltage is $4.2\ \text{mV}$ . The charge carriers are electrons. Calculate the number density of charge carriers.

Solution

Rearrange $V_H = BI/ntq$ :

n = \frac{BI}{V_H t q} = \frac{0.060 \times 20 \times 10^{-3}}{4.2 \times 10^{-3} \times 0.50 \times 10^{-3} \times 1.60 \times 10^{-19}} = 3.6 \times 10^{21}\ \text{m}^{-3}

This is about $10^{7}$ times smaller than for copper, which explains the much larger Hall voltage.

Hall field and drift speed

The slice in the previous example is $6.0\ \text{mm}$ wide. Calculate (a) the electric field strength across the slice in the steady state, (b) the drift speed of the electrons.

Solution

(a) The Hall voltage acts across the width:

E = \frac{V_H}{d} = \frac{4.2 \times 10^{-3}}{6.0 \times 10^{-3}} = 0.70\ \text{V m}^{-1}

(b) In the steady state $qE = Bqv$ , so

v = \frac{E}{B} = \frac{0.70}{0.060} = 12\ \text{m s}^{-1}

(Check with $I = ndtvq$ : $v = 20 \times 10^{-3}/(3.6 \times 10^{21} \times 6.0 \times 10^{-3} \times 0.50 \times 10^{-3} \times 1.60 \times 10^{-19}) = 12\ \text{m s}^{-1}$ .)

Calibrating and using a Hall probe

A Hall probe gives a reading of $36\ \text{mV}$ when placed, face perpendicular to the field, at the centre of a solenoid where $B = 0.090\ \text{T}$ .

(a) Calculate the flux density at a point where the maximum reading is $22\ \text{mV}$ . (b) The probe is returned to the solenoid and rotated so that the normal to its face makes an angle of $60^\circ$ with the field. Calculate the new reading.

Solution

(a) The calibration constant is $k = V_H/B = 36/0.090 = 400\ \text{mV T}^{-1}$ . So

B = \frac{V_H}{k} = \frac{22}{400} = 0.055\ \text{T}

(b) Only the component of $B$ along the normal to the face contributes:

V_H = 36 \cos 60^\circ = 18\ \text{mV}

Changing the slice

A Hall probe gives a Hall voltage of $2.0\ \text{mV}$ . State the new Hall voltage, in the same field, if (a) the current is doubled and the slice is replaced by one of half the thickness, made of the same material, (b) the slice is replaced by one of the same dimensions made of a material with $50$ times as many charge carriers per unit volume, and the current is unchanged.

Solution

$V_H = BI/ntq$ .

(a) $V_H \propto I/t$ . Doubling $I$ doubles $V_H$ ; halving $t$ doubles it again: $V_H = 4 \times 2.0 = 8.0\ \text{mV}$ .

(b) $V_H \propto 1/n$ : $V_H = 2.0/50 = 0.040\ \text{mV}$ . More carriers share the same current, so each drifts more slowly and feels a smaller magnetic force.

Watch out

Using width instead of thickness. In $V_H = BI/ntq$ , $t$ is the thickness, measured parallel to the magnetic field. The width $d$ , across which the voltage is measured, cancels in the derivation. Read diagrams carefully to see which dimension is which.

Watch out

Saying the charges keep moving to the side. The carriers are deflected only until the electric force balances the magnetic force. In the steady state they move straight along the slice and the Hall voltage is constant.

Watch out

Explaining the semiconductor advantage incorrectly. It is not that semiconductors "conduct better" or "have bigger charge". They have a smaller $n$ , so for the same current the carriers drift faster, the magnetic force on each is bigger, and $V_H$ is bigger.

Exam tip

"Explain the origin of the Hall voltage" (3 to 4 marks): charge carriers moving in a magnetic field feel a force (perpendicular to field and current); charge builds up on one side (edge) of the slice; this creates an electric field (p.d.) across the slice; equilibrium when electric force equals magnetic force.
"Show that $V_H = BI/ntq$ ": write $qV_H/d = Bqv$ and $I = nAvq$ with $A = dt$ . Each equation earns a mark; the substitution earns the last.
"Explain why the probe uses a semiconductor rather than a metal": smaller number density $n$ ; so larger drift speed for the same current; so larger Hall voltage.
"Describe how to use a Hall probe to measure $B$ ": face perpendicular to the field; rotate for maximum reading; calibrate in a known field (or zero first); $V_H \propto B$ .
Diagram questions often ask which face becomes positive. Find the force on the carriers with Fleming's left-hand rule, decide which edge they collect on, then give that edge the sign of the carriers.

Summary

Charge carriers drifting through a slice in a magnetic field are pushed sideways by $Bqv$ and collect on one edge.
The separated charge creates an electric field across the slice. Equilibrium when $qV_H/d = Bqv$ .
Using $I = nAvq$ with $A = dt$ : $V_H = BI/ntq$ , where $t$ is the thickness parallel to $B$ .
$V_H$ is larger for a thinner slice, a larger current and a smaller carrier density, so semiconductors are used.
For a given probe and current, $V_H \propto B$ : calibrate once, then read $B$ directly.
Hold the face perpendicular to the field and rotate for the maximum reading; a tilted probe reads $B\cos\alpha$ .

Practice questions

Question

Explain why a Hall voltage is set up across a current-carrying slice placed in a magnetic field, and why it reaches a constant value.
Starting from the forces on a charge carrier, derive $V_H = BI/ntq$ . State the meaning of $t$ .
A semiconductor slice $0.20\ \text{mm}$ thick carries a current of $50\ \text{mA}$ in a field of $0.20\ \text{T}$ perpendicular to its face. The carrier density is $2.0 \times 10^{22}\ \text{m}^{-3}$ and the carriers have charge $e$ . Calculate the Hall voltage.
A metal foil $0.050\ \text{mm}$ thick carries $3.0\ \text{A}$ in a field of $1.2\ \text{T}$ . The Hall voltage is $6.0\ \mu\text{V}$ . Calculate the number density of free electrons.
Explain why Hall probes are made from semiconductors rather than metals.
A calibrated Hall probe has a constant of $0.40\ \text{V T}^{-1}$ . Held with its face perpendicular to the Earth's magnetic field, it reads $18\ \mu\text{V}$ . Calculate the flux density of the Earth's field.
A semiconductor slice is $4.0\ \text{mm}$ wide and $0.80\ \text{mm}$ thick, with $3.0 \times 10^{21}$ electrons per cubic metre. It carries $12\ \text{mA}$ in a field of $0.25\ \text{T}$ perpendicular to its face. Calculate (a) the Hall voltage, (b) the drift speed, (c) the electric field strength across the slice. Show that your answers to (b) and (c) are consistent with the force balance.
A student uses a Hall probe to measure the flux density at different distances from a long straight wire carrying a steady current. Describe how she should take the measurements, and explain two precautions that improve the accuracy.

Answers

The charge carriers move through the magnetic field, so each experiences a magnetic force $Bqv$ perpendicular to its velocity and the field. Carriers collect on one edge, leaving the opposite edge with the opposite charge, so a p.d. (and electric field) develops across the slice. The electric force on each carrier opposes the magnetic force and grows as charge builds up. When the two forces are equal, there is no further sideways deflection and the Hall voltage is constant.
Steady state: $qE = Bqv$ with $E = V_H/d$ , so $V_H = Bvd$ . Current $I = nAvq$ with $A = dt$ , so $v = I/(ndtq)$ . Substitute: $V_H = BId/(ndtq) = BI/ntq$ . $t$ is the thickness of the slice, measured in the direction of the magnetic field.
$V_H = BI/ntq = (0.20 \times 0.050)/(2.0 \times 10^{22} \times 0.20 \times 10^{-3} \times 1.60 \times 10^{-19}) = 0.016\ \text{V}$ ( $16\ \text{mV}$ ).
$n = BI/(V_H t q) = (1.2 \times 3.0)/(6.0 \times 10^{-6} \times 0.050 \times 10^{-3} \times 1.60 \times 10^{-19}) = 7.5 \times 10^{28}\ \text{m}^{-3}$ .
A semiconductor has a much smaller number density of charge carriers. For the same current, the carriers must have a much larger drift speed ( $I = nAvq$ ), so the magnetic force on each is larger and the Hall voltage ( $\propto 1/n$ ) is much larger and measurable.
$B = V_H/k = 18 \times 10^{-6}/0.40 = 4.5 \times 10^{-5}\ \text{T}$ .
(a) $V_H = BI/ntq = (0.25 \times 12 \times 10^{-3})/(3.0 \times 10^{21} \times 0.80 \times 10^{-3} \times 1.60 \times 10^{-19}) = 7.8 \times 10^{-3}\ \text{V}$ . (b) $v = I/(ndtq) = 12 \times 10^{-3}/(3.0 \times 10^{21} \times 4.0 \times 10^{-3} \times 0.80 \times 10^{-3} \times 1.60 \times 10^{-19}) = 7.8\ \text{m s}^{-1}$ . (c) $E = V_H/d = 7.8 \times 10^{-3}/4.0 \times 10^{-3} = 1.95\ \text{V m}^{-1}$ . Consistency: in the steady state $E = Bv = 0.25 \times 7.81 = 1.95\ \text{V m}^{-1}$ , matching (c).
Keep the current constant (check with an ammeter). Measure the distance from the centre of the wire to the sensing element, and place the probe with its face perpendicular to the field lines (circles around the wire), rotating it to the maximum reading at each position. Take readings for at least six distances and plot $B$ against $1/r$ . Precautions: (i) zero the probe with the current off, or take readings with the current in both directions and average, to remove the effect of the Earth's field; (ii) keep the probe orientation fixed with a clamp and set square, because a tilted probe reads only the component $B\cos\alpha$ .

Setting the scene

Where the Hall voltage comes from

Deriving the Hall voltage equation

What makes a large Hall voltage

The Hall probe

Orientation matters

Worked examples

Practice questions

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