Chapter 3

Physics of detection and identification

How metal targets respond to alternating magnetic fields, and what a frequency sweep adds.

This chapter summarises the physical principles that the design relies on. All curves are computed in the browser from the equations given in the text.

3.1Electromagnetic induction

A transmit coil carrying alternating current produces a magnetic field. The field induces eddy currents in any conductor nearby, and the eddy currents produce a secondary field that a receive coil detects. At the frequencies used here the wavelength is kilometres long, so all coupling takes place in the magnetic near field; no radio waves are involved.

Table 3.1.Principal types of metal detector.

Type	Frequency	Principle	Typical strength
VLF induction balance	3 to 30 kHz	Continuous tone; amplitude and phase of the secondary field are measured	Discrimination between metals
Pulse induction	pulsed, µs decay	Transmitter switched off; decay of eddy currents is timed	Depth, tolerance of mineralised ground
Beat frequency oscillator	about 100 kHz	Metal changes coil inductance and shifts an oscillator	Simplicity

Low frequencies are used because a field entering a conductor decays over the skin depth

$$\delta = \sqrt{\frac{2\rho}{\omega\mu}},$$

which in copper is about 0.65 mm at 10 kHz and about 2 µm at 1 GHz. Soil is a weak conductor and follows the same law, so induction at kilohertz frequencies penetrates the ground while radio frequencies are attenuated near the surface. A VLF detector already has the structure of a software-defined receiver. It transmits a tone, multiplies the received signal by the sine and cosine of the reference (I/Q demodulation), and reads amplitude and phase. The design in Chapter 5 implements this with an audio codec.

3.2Response of a conducting target

A simple target, such as a ring, can be modelled as a loop with inductance $L_t$ and resistance $R_t$. Its response to a field at angular frequency $\omega$ is

$$H(\omega) = \frac{j\omega\tau}{1 + j\omega\tau}, \qquad \tau = \frac{L_t}{R_t} \sim \mu\sigma\ell^2 ,$$

where $\ell$ is a characteristic size. This is a relaxation, not a resonance: the magnitude increases monotonically with frequency. The quadrature (90°) component has a maximum at $\omega = 1/\tau$, and this maximum is how the time constant is measured in practice.

Figure 3.1.Response of a single-time-constant target. The magnitude (dashed) has no maximum; the quadrature component peaks at $\omega\tau = 1$.

Because $\tau \sim \mu\sigma\ell^2$ contains both material and size, a large aluminium object and a smaller copper object can have the same time constant. This is the basic ambiguity of metal identification by induction.

3.3The conducting, permeable sphere

For a solid sphere of radius $a$, conductivity $\sigma$ and relative permeability $\mu_r$ in a uniform field, the induced dipole moment is known exactly [R16]:

$$\frac{M}{2\pi a^3} = \frac{2\mu_r\,(x\coth x - 1) - (1 + x^2 - x\coth x)}{\mu_r\,(x\coth x - 1) + (1 + x^2 - x\coth x)},\qquad x = a\sqrt{j\omega\mu_0\mu_r\sigma}.$$

The expression was checked against three limits: at zero frequency it gives the magnetostatic value $2(\mu_r-1)/(\mu_r+2)$; at high frequency it tends to $-1$ (a perfect diamagnet); and at low frequency for a non-magnetic sphere it reduces to the eddy-current result $-x^2/10$. All three agree to numerical precision.

Figure 3.2.In-phase and quadrature response of five spheres, normalised by volume, from 10 Hz to 1 MHz. The shaded band is the range covered by the codec at 48 kHz sampling. The dashed aluminium curve (radius 6.3 mm) coincides with the copper curve (radius 5 mm).

For non-magnetic spheres the quadrature peak lies at $f_\text{peak} \approx 1.85/(\mu_0\sigma a^2)$. This is about 1.0 kHz for a copper sphere of 5 mm radius, 1.6 kHz for aluminium and 41 kHz for stainless steel. An aluminium sphere of 6.3 mm radius has the same $\sigma a^2$ as the 5 mm copper sphere, so their normalised spectra are identical; only the absolute amplitude differs, by a factor of $(6.3/5)^3 \approx 2$. A steel sphere ($\mu_r \approx 100$) remains on the ferrous side of the response over the whole audio band; its in-phase component changes sign only near 280 kHz.

3.4One frequency compared with many

A detector working at one frequency measures one complex number per position: an amplitude, which combines size with depth, and a phase. In the complex plane every target traces a curve as frequency increases, and a single-frequency measurement is one point on that curve.

Figure 3.3.Response loci in the complex plane. All non-magnetic solid spheres lie on one common curve; conductivity and size only determine where on it a given frequency falls. A ring follows a semicircle. Points mark a 10 kHz measurement.

A frequency sweep records the whole curve. Its shape carries information that a single point cannot. The clearest example is geometry. A ring has one eddy-current path and one time constant. A solid object has many superposed modes, which widens its response.

Figure 3.4.Quadrature response of a ring and of a solid sphere, each shifted so that its peak is at the centre. The sphere's response is lower and about 40% wider (1.56 against 1.13 decades at half height).

Table 3.2.What one frequency and a sweep can determine.

Question	One frequency	Sweep
Ferrous or non-ferrous	Yes (phase)	Yes
Product $\sigma a^2$	Only for simple shapes	Yes, by fitting the full curve
Ring, solid or flat object	No	Yes, from curve width and shape
Mineralised ground	Suppressed at one phase, together with targets at that phase	Separable by spectral shape
Two objects under the coil	No	Partly
Copper or aluminium, size unknown	No	No; requires size from geometry (Chapter 7)

Magnetically viscous soil, which contains very fine magnetic grains, shows a quadrature response that is nearly constant with frequency. Its in-phase response falls roughly linearly with the logarithm of frequency. No metallic target behaves like this, so with a sweep the ground response can be fitted and removed instead of being suppressed at one phase.

3.5Dipole model and distance dependence

A small target in the field of a transmit coil becomes a magnetic dipole. Its strength and direction are set by a symmetric 3 × 3 polarizability tensor $\mathbf M(\omega)$. The voltage induced in a receive coil is

$$V(\omega) \propto \mathbf h_\text{rx}^{\mathsf T}(\mathbf r)\,\mathbf M(\omega)\,\mathbf h_\text{tx}(\mathbf r),$$

where $\mathbf h_\text{tx}$ is the transmitter's field at the target and $\mathbf h_\text{rx}$ the receiver's sensitivity there. Both fall off as $1/r^3$, so for a small target

$$|V| \propto \frac{|\mathbf M|}{r^6}.$$

Doubling the depth reduces the signal by a factor of 64. In the target's own axes the tensor is diagonal:

$$\mathbf M(\omega) = \mathbf R^{\mathsf T}\,\mathrm{diag}\big(\beta_1(\omega), \beta_2(\omega), \beta_3(\omega)\big)\,\mathbf R.$$

The three curves $\beta_i(\omega)$ describe the object independently of its orientation. A sphere gives three equal curves; elongated and flat objects give unequal ones. Recovering the $\beta_i$ is the central measurement task of the project. Chapter 5 shows which coil arrangement makes it possible.

Question	One frequency	Sweep
Ferrous or non-ferrous	Yes (phase)	Yes
Product \(\sigma a^2\)	Only for simple shapes	Yes, by fitting the full curve
Ring, solid or flat object	No	Yes, from curve width and shape
Mineralised ground	Suppressed at one phase, together with targets at that phase	Separable by spectral shape
Two objects under the coil	No	Partly
Copper or aluminium, size unknown	No	No; requires size from geometry (Chapter 7)