Moving charges inside a wire
So far the charges have flown through empty space. The charges that carry a current in a metal also move, so a magnetic field pushes them too. The chapter Electric Current and Resistance describes them. The free charges, or charge carriers, drift along the conductor with a small average velocity \(v_d\), the drift velocity, and the current is \(I = nqAv_d\). Here \(n\) is the number of carriers per cubic metre and \(A\) the cross-section.
Take a flat strip of width \(w\) and thickness \(d\), carrying a current \(I\) along its length. A field \(\vec B\) crosses the strip at right angles to its flat face. Each carrier feels \(q\,\vec v_d\times\vec B\) across the width of the strip. The carriers are pushed towards one edge, so charge piles up on that face, and the opposite face is left with the opposite charge.
The piled-up charge makes an electric field \(E_H\) across the strip, and that field pushes the carriers back. Charge keeps piling up until the electric force balances the magnetic one: \(qE_H = qv_dB\). Then the carriers drift straight along again, just as the right speed passes a velocity selector. The whole balance is reached in a tiny fraction of a second.
A current flows to the right along the strip, in a field pointing into the page. The current is carried by electrons, drifting to the left. Which face of the strip becomes negative?
In the figure above the magnetic force \(F_B\) on each carrier stays the same. The electric force \(F_E\) of the piled-up charge starts at zero and grows until it matches \(F_B\). Their ratio, read in the lower box, climbs to \(1.00\). The build-up is slowed enormously in the figure; in a real strip it is over almost at once.
The Hall voltage
In balance the field across the strip is \(E_H = v_dB\). Across the width \(w\) it gives a potential difference, the Hall voltage, \(V_H = E_Hw = v_dBw\). Put in \(v_d = I/(nqwd)\), since the cross-section is \(A = wd\). The width cancels and \(V_H = IB/(nqd)\). This transverse voltage is the Hall effect.
A copper strip \(1\,\text{cm}\) wide and \(1\,\text{mm}\) thick carries \(10\,\text{A}\) in a field of \(1\,\text{T}\) across its face. Copper has \(8.5\times10^{28}\) free electrons per cubic metre. Find the drift speed and the Hall voltage.
\(v_d = \dfrac{I}{neA} = \dfrac{10}{8.5\times10^{28}\times1.6\times10^{-19}\times10^{-5}} = 7.4\times10^{-5}\,\text{m/s}\), less than a tenth of a millimetre per second.
\(V_H = v_dBw = 7.4\times10^{-5}\times1\times0.01 = 7.4\times10^{-7}\,\text{V}\). The same from \(IB/(ned) = 10/(8.5\times10^{28}\times1.6\times10^{-19}\times10^{-3})\).
Less than a microvolt, even with ten amperes in a whole tesla, because copper has an enormous number of carriers.
Values used: \(e = 1.6\times10^{-19}\,\text{C}\), \(n = 8.5\times10^{28}\,\text{m}^{-3}\) for copper.
The tiny result has a clear cause. \(V_H = IB/(nqd)\) is inversely proportional to \(n\). A metal has so many carriers that each one drifts slowly, and the balancing field is weak. A semiconductor has far fewer carriers, often around \(10^{21}\,\text{m}^{-3}\). The unit on Modern Physics describes these materials. With few carriers, each must drift fast to carry the same current, and the Hall voltage is large.
A copper strip \(0.5\,\text{mm}\) thick carries \(8\,\text{A}\) in a field of \(0.5\,\text{T}\) across its face. With \(n = 8.5\times10^{28}\,\text{m}^{-3}\), find the Hall voltage in microvolts.
A metal strip \(0.1\,\text{mm}\) thick carries \(2\,\text{A}\) in a field of \(0.4\,\text{T}\), and shows a Hall voltage of \(0.5\,\mu\text{V}\). Find the number of carriers per cubic metre, in units of \(10^{29}\), assuming each carries charge \(e\).
Which charges carry the current?
A current to the right could be positive charges moving right or negative charges moving left. Both push the same current, and an ordinary meter cannot tell them apart. The Hall effect can. Reversing the sign of the carriers also reverses their drift velocity, so \(q\,\vec v_d\times\vec B\) stays the same. Carriers of either sign are pushed to the same face. The sign of the charge on that face is then the sign of the carriers.
Choose the second chip of the figure above. The positive carriers drift to the right and are pushed up, as the electrons were. Now the top face turns positive instead of negative. The Hall voltage has the opposite sign. In copper, silver and most metals the face shows that the carriers are negative: they are electrons. In a few metals, zinc among them, the sign comes out positive; explaining that needs the theory of solids.
A strip carries a current to the right in a field into the page, and the current is carried by positive charges. Which face of the strip becomes positive?
Edwin Hall was a graduate student at Johns Hopkins University in Baltimore. He asked whether a magnet acts on the current itself or only on the wire. In 1879 he passed a current through a thin strip of gold leaf between the poles of an electromagnet. Across the width of the strip he found a tiny steady voltage. It was the first direct sign that the magnetic force acts on the moving charge inside the metal.
Many new energy-saving ceiling fans in India use a brushless motor. Its coils must be switched on and off in turn as the magnets of the rotor go past. A small semiconductor Hall sensor beside the rotor reads the field of the nearest magnet. Its Hall voltage changes sign as a north pole gives way to a south pole, and the fan's circuit switches the coils. A sensor with \(n = 10^{21}\,\text{m}^{-3}\), \(0.5\,\text{mm}\) thick, carrying \(1\,\text{mA}\) in \(0.1\,\text{T}\) gives \(1.25\,\text{mV}\). That is easy to read.
A Hall sensor for measuring magnetic fields is made from a semiconductor, not from copper. Explain why.
The Hall voltage is \(V_H = IB/(nqd)\), inversely proportional to the number of carriers per unit volume. Copper has about \(10^{29}\) carriers per cubic metre, so even a large current in a strong field gives less than a microvolt. A semiconductor has far fewer carriers, so the same current and field give a Hall voltage many thousands of times larger. That voltage is large enough to measure easily and is proportional to \(B\).
- Taking the Hall voltage across the thickness instead of the width.\(V_H\) appears across the width \(w\), perpendicular to both the current and \(\vec B\). The thickness \(d\) is along \(\vec B\).
- Expecting positive and negative carriers to go to opposite faces.Both the charge and the drift direction reverse, so the force is the same; the face's sign reveals the carriers.
- Using the width in \(IB/(nqd)\).The width cancels; \(d\) is the thickness measured along the field.