Calculate power supply on distance
20 gauge wire has a resistance of about \$ 33.31m\Omega / m \$. For a 20 meter cable, it's effectively a resistor \$33.31\frac{\Omega}{m} \cdot 20m \approx 1 \Omega \$. Your circuit is this:



You want 30A through the motor, and since this is a series circuit, that means 30A through R1 and R2, also. The voltage drop over a resistor is given by Ohm's law: voltage equals current times resistance:

\$ V_{R1} = 30A \cdot 1\Omega = 30V \$

So, you lose 30V in R1, and another 30V in R2, so you will need 60V plus whatever voltage is required at the motor to get 30A. Worse, power equals the product of voltage and current. We know the voltage across R1 is 30V, and current is 30A, so power lost in the wire is:

\$ P_{R1} = 30A \cdot 30 V = 900 W \$

For comparison, a typical electric heater is around 1000W. You have two wires, so your total losses are 1800W. It's quite likely you will trip a circuit breaker if you can magically prevent the wires from bursting into flames, and we haven't even powered the motor yet.

There are two obvious solutions:


make the wires shorter
make the wires fatter


If you can't do either of those, there's a less obvious solution:


keep the power the same by raising the voltage and reducing the current


This is the solution the electric company uses to avoid huge losses in power transmission. If you combine the two previous equations, you can see that the power through a fixed resistance can be calculated through the current alone:

\$ P = I^2 R \$

12V at 30A is 360W. 360V at 1A is also 360W, and in theory, capable of producing the same mechanical power at your motor. But, the losses in your wires will be much less, by minimizing the \$I^2\$ term above.

To do this, look for a motor that operates at a higher voltage, or put a mechanism for converting the voltage near the motor.