Part $a$ of the figure shows a heterodyne metal detector being used. As part $b$ of the figure illustrates, this device utilizes two capacitor/ inductor oscillator circuits, A and B. Each produces its own resonant frequency, $f_{0 \mathrm{A}}=1 /\left[2 \pi\left(L_{\mathrm{A}} C\right)^{1 / 2}\right]$ and $f_{0 \mathrm{B}}=1 /\left[2 \pi\left(L_{\mathrm{B}} C\right)^{1 / 2}\right]$. Any difference between these frequencies is detected through earphones as a beat frequency $\mid f_{0 \mathrm{B}}-$ $f_{0 A} \mid .$ In the absence of any nearby metal object, the inductances $L_{\mathrm{A}}$ and $L_{\mathrm{B}}$ are identical. When inductor $\mathrm{B}$ (the search coil) comes near a piece of metal, the inductance $L_{\mathrm{B}}$ increases, the corresponding oscillator frequency $f_{\mathrm{oB}}$ decreases, and a beat frequency is heard. Suppose that initially each inductor is adjusted so that $L_{\mathrm{B}}=L_{\mathrm{A}},$ and each oscillator has a resonant frequency of $855.5 \mathrm{kHz}$. Assuming that the inductance of search coil $\mathrm{B}$ increases by $1.000 \%$ due to a nearby piece of metal, determine the beat frequency heard through the earphones.