High-current DC power supply in constant current mode with inductive load
10A or 100A constant DC current to a milliohm range load is possible without heavy losses by using a switchmode circuit. The basic idea is to utilize the induction. The inductor may be your load or an extra inductor for the current regulation only. 

How the inductor works: When a dc voltage is applied to a inductor its current starts to grow at rate voltage/inductance. 

If the inductor has a parallel back emf suppression diode, the dc voltage can be disconnected without a spark. After the disconnection the inductor current diminishes at rate Vd/inductance. Vd means the fordward voltade drop of the diode. By switching the dc on off at high enough frequency, the current swings between two values just like the thermostat makes the temperature to swing between the upper and the lower limit. Here is the principle of the circuit:



Rs is the current measuring shunt (milliohms). The Osc has a schmitt-trigger that turns the fet on and off +proper high speed drive circuit for the fet. With no more in the Osc the system is a free running oscillator. 

A switchmode supply control IC surely is adaptable to this for PWM control and  wanted operating frequency. The free running system can be moved to higher frequency by narrowing the current hysteresis.

A milliohm range resistive load can be inserted in series with the inductor. If the inductor generates an extra emf due an external magnetic field, the +Vs must win it to keep the current. If the inductive load is an electomagnet that moves iron, high external emf pulses are expectable.

D1 must be a fast turn off and low voltade drop type and be rated for the reverse voltage Vs + the external emf. I have drawn a schottky diode just for remembering the speed and low drop.

Development ideas:  

1) 100 amperes sounds high. It surely would be useful to investigate if there  is available some modern switching components for high current motors.

2) the dissipation in the fet grows to enormous if there is no inductance. To be in safe some protection scheme that sees the simultaneous high fet current and voltage should be added.

3) the dissipation in D1 can be intolerably high. An active switch can have much lower fordward voltage drop. 50% smaller drop means 50% smaller dissipation. This is important because the current runs most of the time through the diode, the fet only gives a new kick when it's needed.