Understanding the meaning of amplitude in FFT
The continuous-time Fourier transform of a function $f(t)$ is in essence an integration of $f(t)$ multiplied with a complex exponential kernel:
$$ F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt \tag{1}$$
Since the exponential function is unitless, the unit of the Fourier integral will be the multiplication of the units of the function $f(t)$ and the differential $dt$.
Assuming that the function $f(t)$ had a unit of micro Tesla, and its argument $t$ is time (in seconds), then the unit of $dt$ will be seconds. As a consequence, the unit of the Fourier transform, $F(\omega)$, will be  micro Tesla  second  $$\mu T \cdot s \tag{2}$$
However, what you actually compute is the discrete-time Fourier transform, $F(e^{j\omega})$, of the samples $f[n]= f(nT_s)$ of the function $f(t)$, via the summation: $$F(e^{j\omega}) = \sum_{n=-\infty}^{\infty} f[n] e^{-j\omega n} \tag{3}$$ where $T_s$ is the sampling period in seconds.
Furthermore, instead of the continuous-argument function $F(e^{j\omega})$, you will compute its samples $F[k]$
$$ F[k] = \sum_{n=0}^{N-1} f[n] e^{-j \frac{2\pi}{N} n k} \tag{4}$$
through a DFT (discrete Fourier transform) of the samples $f[n]$ of length $N$, possibly implemented with an FFT algorithm.
The unit of the samples $f[n]$ is the same as the unit of $f(t)$, making the unit of $F(e^{j\omega})$  and $F[k]$ as micro Tesla. Therefore the unit of the FFT samples $F[k]$ of $F(e^{j\omega})$ will be micro Tesla.
Note that there is an (implicit) amplitude scaling by $1/T_s$ in the computed DFT samples $F[k]$, and when you want to display the continuous-time Fourier transform $F(\omega)$ from the samples $F[k]$, you multiply them with $T_s$, which corrects not only the amplitude scaling, but also the unit of it, by making it micro Tesla second as in (2).