At the battery, say it is at 4V, then 30W will require (watts = volts x current) 30/4 = 7.5A. There are small converter inefficiences so it will probably draw around 8A.
On the coil side, the 0.1 ohm coil will get 30W of power. Power = volts squared/R or Power = current squared x R, that gives us 1.73V and 17.3A to the 0.1 ohm coil. But as the coil heats up its resistance will approximately double, so the volts and amps will change in unison.
That is what a DC-DC converter can do, that is, to take voltage and current from a battery and convert it to a different voltage and current to power the coil. Voltage x current = Watts. Here we have 30 watts taken from the battery and 30 watts delivered to the coil, but in different numbers of volts and amps. The battery life is unaffected, the battery is simply driving a 30 watt load, it doesn't know there's a 0.1 ohm coil or whatever on the other side of the DC-DC converter.