Coil output from variable magnetic field
The open circuit output voltage is \$N\frac{d\Phi}{dt}\$ so, if you can position the coil so that it receives the largest amount of \$\frac{d\Phi}{dt}\$ you will get the largest peak-to-peak output voltage. Clearly, if the N turns have a cross sectional area that allows more flux (\$\Phi\$) to enter then you also get a larger output voltage. More cross section usually means more voltage. More N means more open circuit voltage. 

A higher operating frequency can sometimes mean a higher \$\frac{d\Phi}{dt}\$ except in a regular transformer where the magnetization inductive reactance rises linearly with frequency and \$\frac{d\Phi}{dt}\$ remains constant for a given excitation voltage.

If the coil is loaded so that a significant current is drawn then things get more complex unless the coupling is close to 100% (as in a regular transformer). In non 100% coupled scenarios, you need to try and work out what the mutual inductances are so that you can estimate the output voltages when current is being drawn. 

However, you can "know" k (the coupling factor) by studying the open circuit voltage and this can help you find the mutual inductances. It rapidly becomes a non-trivial exercise though.