Electromagnetic crack detectors and metal detectors
It is possible to detect cracks in metallic objects as follows. If the part to be examined is placed in the vicinity of a coil fed with alternating current, the inductance measured at the coil terminals is lowest when there are no cracks. Such instruments can detect cracks only. 10 micrometers deep. The coil forms part of a resonant circuit. $^{*}$ Metal detectors operate similarly.
Consider the following simpler situation. A single-layer close-wound solenoid has a length $l$, a radius $a$, and $N$ turns. Let us calculate how its impedance changes when one introduces into the solenoid a thin brass tube of wall thickness $b$.
When an alternating current flows in the solenoid, the changing magnetic flux induces a current in the tube, which thus acts as a secondary winding. According to Lenz's law, the induced current tends to cancel $d \Phi / d r$, and hence $j \omega \Phi$, and hence $\Phi$. The presence of the tube thus reduces the inductance at the solenoid terminals. The effective inductance of the solenoid decreases when the resistance of the tube decreases.
We disregard the skin effect (Sec. 29.1) in the tube and the stray capacitance of the coil. Also, we set $l \gg a$ so as to disregard end effects. The coefficient of coupling is nearly equal to unity.
(a) Calculate the resistance $R_{1}$ of the winding of the solenoid. Set the conductivity of copper equal to $\sigma_{e}$.
(b) Calculate the impedance $Z_{t}$ of the solenoid without the brass tube.
(c) Calculate the resistance $R_{2}$ of the brass tube in the azimuthal direction. Set its radius equal to $a$, and call its conductivity $\sigma_{b}$,
(d) Calculate its inductance $L_{2}$ and impedance $Z_{2}$
(e) Now calculate the impedance at the solenoid terminals with the tube in place.
(f) Calculate impedances, without and with the brass tube, when $N=1000, \quad l=200$ millimeters, $a=20.0$ millimeters, $b=0.5$ millimeter $f=1000$ hertz, $\quad \sigma_{e}=5.8 \times 10^{7}$ siemens/meter, $\quad \sigma_{b}=1.6 \times 10^{7}$ siemens $/$ meter. Note how the presence of the tube increases $R$ (more dissipation) and decreases $L$ (less flux).