# Why does the V/Hz ratio of a motor affect the magnetic field strength?

When the frequency is increased whilst keeping the voltage constant, the magnetic field strength is going to weaken... I don't understand why and what actually happens that causes the field strength to drop.

If anybody has an information on this I'd appreciate some clarity on the matter.

• A hint: consider how that affects the current, given that motors are inductive – Hearth Oct 14 '19 at 16:13
• Moving rotor generates almost the same voltage as driver (BEMF) and thus with no load current is only 5%~10% of load current. Surge current at stall is V+/ DCR of coil can be 5~10x due to DCR/Zcoil(f) at full loaded RPM. Thus to flatten torque = current vs RPM, V is increased with f. – Tony Stewart EE75 Oct 14 '19 at 16:38
• current thru an inductor, which your coils are, is I = 1/L * integral(V*dT). This means faster frequency gives a smaller dT, and the peak current is reduced. – analogsystemsrf Oct 14 '19 at 17:10

Consider the equivalent circuit for one phase of an induction motor shown below. The current, $$\Im\$$ in the magnetizing inductance $$\Xm\$$ determines the stator magnetic field strength. For a simple approximation, we can assume that $$\Xs = 0\$$ and everything else in the circuit except $$\Xm\$$ is infinite. We can then say that $$\Im = \frac{V}{Xm}\$$
Since $$\Xm = 2*Pi*f*Lm\$$, we can say that $$Im = \frac{V}{f}*\frac{1}{(2*Pi*Lm)}$$.