4.1.9 Applications

Send a current $I$ along a flat strip of thickness $t$ and width $w$ placed in a magnetic field $B$ normal to its face (Figure 4.1.13). E. H. Hall discovered in 1879 what happens. Each carrier, drifting along the strip with velocity $v$, feels the Lorentz force $qv\times B$ sideways, across the width. The carriers are deflected towards one long edge, charge piles up there (and the opposite charge is left on the other edge), and a transverse electric field $E_H$, the Hall field, builds up across the strip. The pile-up stops when the electric force on a carrier balances the magnetic one:

The current is $I=nqvwt$ (Section 4.1.2, with $n$ the carrier density and $wt$ the cross-section), so $v=I/(nqwt)$, and the voltage between the two edges, the Hall voltage, is

Watch the edge charge build up until $q{E}_H$ balances $qvB$, switch between electrons and holes, then pick a semiconductor.

Three consequences make this simple effect one of the most used in electrical engineering.

• The sign of $V_H$ gives the type of the carriers, and $|V_H|$ gives their density $n$. Electrons moving against $I$ end on the same edge as positive carriers moving with it (both $q$ and $v$ change sign in $qv\times B$), so the edge that goes positive tells which kind of carrier is present. This is how the carrier type and density of a semiconductor are measured.
• $V_H\propto 1/n$. In copper, with $n=8.5\times 10^{28}$ m${}^{-3}$, a strip of $100$ μm carrying $1$ A in $1$ T gives only $0.73$ μV (Worked example 4.1.2). Hall probes are therefore made of thin semiconductors (InSb, GaAs, Si) with $n\sim 10^{21}$ to $10^{22}$ m${}^{-3}$, a million times fewer carriers and a million times more voltage: millivolts from a few milliamperes.
• $V_H\propto B$ at fixed $I$. The element is a magnetometer, and when $B$ is produced by a current to be measured it becomes a contactless current sensor: the Hall element in the gap of the clamp meter of Application 4.1.3, the current sensors of motor drives and battery monitors. The same elements detect the rotor magnets of brushless DC motors and the teeth of a gear wheel, giving position and speed without contact.

In a wire the moving charges are the carriers, and summing $qv\times B$ over a straight length $l$ gave the wire form of the Lorentz force, Eq. (4.1.5):

The current, not the carrier sign, fixes the direction. A $1$ m wire with $10$ A in $0.5$ T feels $5$ N. Put such a wire on a shaft and let it turn, and you have a motor.

The armature loop. Figure 4.1.14 shows the simplest DC motor: a rectangular loop of length $l$ and width $2a$ on a shaft, in the field $B$ between the pole pieces N and S. The current runs opposite ways along the two long sides, so the two forces $F=IlB$ are equal and opposite: no net force on the loop, but a torque about the shaft. With the loop plane parallel to $B$ the lever arm of each force is $a$ and the torque is $2aF$. In general, with $\alpha$ the angle between $B$ and the normal of the loop, the lever arm is $a\sin \alpha$, and for a coil of $N$ turns

The commutator. As the loop turns, $\sin \alpha$ changes sign every half revolution and the torque would reverse: the loop would simply swing to the position $\alpha =0$ and stop. The split ring and the fixed brushes reverse $I$ in the loop exactly when $\sin \alpha$ changes sign, so the torque is rectified to $T_{\max }|\sin \alpha |$ and always drives the same way. Its mean over a revolution is $(2/\pi )T_{\max }$ (Worked example 4.1.4). A real armature carries many coils spaced round the rotor, each commutated at its own moment, so the total torque stays close to $T_{\max }$ instead of pulsating. A brushless motor replaces the commutator by electronic switches timed by Hall sensors (Application 4.1.1), but uses the same $Il\times B$.

Run the coil with the commutator off in Free run, then switch it on and watch the torque keep one sign.

The $1/\rho$ field of the straight wire, Eq. (4.1.21), is the working formula of power engineering: $B={\mu }_{0}I/(2\pi \rho )$ with ${\mu }_0/2\pi =2\times 10^{-7}$ H/m. It is used in two opposite ways, to find the field of a known current and to find a current from its field.

Under an overhead line. One conductor of a $400$ kV line, $15$ m above the ground, carrying $1$ kA, produces at the ground (Figure 4.1.15a and Worked example 4.1.3)

which is below the Earth's own static field of $25$ to $65$ μT at the surface, and about $1/15$ of the ICNIRP 2010 public reference level of $200$ μT at $50$ Hz. On a balanced three-phase line the three currents add to zero at every instant, so the three $1/\rho$ terms cancel and the resultant falls off as $1/{\rho }^2$ (Exercise 4.1.11 shows this for a go-and-return pair): a few microteslas under a real line. The single-conductor value is an upper bound.

Reading a current from its field. Ampère's formula, ${\oint }_{C}H\cdot dl=I_{\mathrm{enc}}$, says that the circulation of $H$ round any closed loop equals the current through it. This is Problem 4.1.4 used backwards: on a circle of radius $\rho$ round a wire,

A clamp meter (Figure 4.1.15b) closes a soft-magnetic ring round one conductor. The iron gathers the flux of the wire's field, a winding on the ring reads an alternating current as a current transformer, and a Hall element in a sawn gap reads direct current too. Its signal depends only on the current inside the loop: the wire may sit anywhere in the jaw, and a neighbouring cable outside the jaw adds nothing, because its contributions to the circulation cancel round the loop (the mistake box of Section 4.1.5). Clamp the meter round a two-core cable carrying the go and the return current and it reads zero. In Worked example 4.1.3 a ring of mean radius $1.5$ cm round a $20$ A cable sees $H=212$ A/m, and $H\times 2\pi \rho =20$ A whatever the position of the cable.

A Rogowski coil (Figure 4.1.16) does the same without iron: a long flexible solenoid of $n$ turns per metre and cross-section $A$ is closed into a loop $C$ round the conductor. Each turn links the flux ${\mu }_{0}H\cdot a_{t}A$ of the local field along the tangent $a_t$ of the loop, and the total flux linkage is

by Ampère's law, whatever the shape of the loop and the position of the conductor (Exercise 4.1.19). The coil contains no iron, so it never saturates: the same coil reads $10$ A and a $100$ kA fault current. So a current of $1$ kA in a busbar is measured without touching it, and the same $1/\rho$ law sets the clearance: $1$ kA at $30$ cm gives $B=2\times 10^{-7}\times 10^3/0.3=0.67$ mT.

A flat spiral wound from the radius $b$ to the radius $2b$ carries its current in the $a_{\varphi }$ direction on an annulus (Figure 4.1.17). Seen from more than a few turn-pitches away it is the sheet current of Problem 4.1.8: a surface current density $J_S=K({\rho }^{\prime })a_{\varphi }$ on the ring $b\le {\rho }^{\prime }\le 2b$, with $K=NI/b$ for $N$ turns spread uniformly over the width $b$.

The field at the centre is found by cutting the sheet into rings. The ring between ${\rho }^{\prime }$ and ${\rho }^{\prime }+d{\rho }^{\prime }$ carries $dI=Kd{\rho }^{\prime }$ and, by Eq. (4.1.22) with $h=0$, contributes $d{H}_z=dI/2{\rho }^{\prime }$ at the centre. Hence, for $N$ turns spread uniformly from $b$ to $2b$,

Wireless-charging pads (the Qi standard, $100$ to $200$ kHz), induction hobs (the bottom of the pan is the "receiver") and NFC antennas ($13.56$ MHz) are all such coils. Because $d{H}_z=dI/2{\rho }^{\prime }$, the inner turns count most at the centre, while the outer turns add less there but extend the field further along the axis (Exercise 4.1.20): a charging pad wants a flat field over the receiver, so its turns are spread outward. At $100$ to $200$ kHz a pad coil runs with a few amperes and makes a few hundred microteslas, enough to transfer several watts across a few millimetres.