Bose-Einstein condensate: anti-Helmholtz coils and temperature dependence, if one is observed To get a Bose-Einstein condensate anti-Helmholtz coils are used to hold the BEC together.
I am looking for a relationship between the strength of the magnetic field of the coils and the temperature at which a given substance becomes a BEC. (Just to avoid unnecessary discussion, I am aware that the anti-Helmholtz coil at the centre has no magnetic potential).
 A: 
To get a Bose-Einstein condensate anti-Helmholtz coils are used to
hold the BEC together.

Not always. You can have purely optical traps, or magneto-optical hybrid traps.

the strength of the magnetic field of the coils and the temperature at
which a given substance becomes a BEC

Changing the strength of the field and hence of the gradient just causes the trap volume to increase and decrease. This compresses and relaxes the atomic cloud. The temperature does change, sure, but the phase space density stays fixed. So you cannot get a BEC just by changing the field strength. You need a dissipation mechanism (e.g. evaporative cooling) to lose entropy and increase phase space density.
EDIT
In a harmonic trap, the critical temperature for condensation $T_{\text{c}}$ is:
$$ T_{\text{c}} = 0.94 \frac{\hbar\bar\omega}{k_{\text{B}}}N^{1/3}, $$ where $N$ is the number of atoms and $\bar\omega$ is the geometric mean of the $x$, $,y$, and $z$ trapping frequencies.
If you can figure out what trapping frequency your magnetic trap has, almost certainly you'll get some dependence $\omega \propto B^{(\text{some power})}$. More field causes a tighter trap which brings the critical temperature up.
The problem with the above is: if you just compress the trap, then both your current temperature $T$ and the critical one $ T_{\text{c}}$ will go up. To get a BEC, you need $T$ to go down faster than $ T_{\text{c}}$ goes down. Hence the need for dissipative processes.
