Motor loaded at percent %
The equation should be
$$I= \frac{P}{V*\sqrt3 *Pf *Eff}$$
Where P is the output power in Watts, V is the line-to-line voltage, Pf is the power factor and Eff is the efficiency.
What is meant by 50% load is that the motor output power is 50 percent of full load output power.
Both power factor and efficiency will change a bit when you go from full load to 75 percent to 50 percent to 25 percent. So the equation is still valid, but you need to use the Eff and Pf specified for the reduced load condition. I don't know if there is a simple way to calculate the new Eff and Pf.
To a first approximation, current and power will scale together linearly. So at half power, the motor current will be half of the full load current. But if you want a more exact calculation you will need to use Eff(50) and Pf(50).