RMS Current in an AC Generator with Doubled Rotational Speed

[Terocamo]

Homework Statement

The output voltage of an a.c. generator is
\emph{V}=\emph{NBA}\omegasin(\omegat).
When the voltage is connected across an inductor, the rms current in the inductor is found to be 10mA. What will Be the rms current if the rotating speed of the coil inside the generator is doubled?

The Attempt at a Solution

I try to rearrange the equation, what i think is that the reactance of the inductor is constant so that if
Xc=\frac{NBA\omega}{\sqrt{2}Irms} (strangely i cannot get the root 2 shown in the faction)
Then \omega\proptoI_rms
So if the rotational speed doubles the Irms also doubled.
But however the answer is the reactance change and the rms current remains at 10mA.
I don't understand how these kind of question can be solved cause there isn't any constant I can make use of in the question...

 

[rl.bhat]

Vmax = NABω and XL = ωL. Imax = Vmax/ωL = NAB/L.

When When the rotating speed doubles, Vmax = NAB(2ω) and XL= 2ωL

So Imax remains the same. Same is true for Irms.

 

[Terocamo]

I don't understand. How is the reactance doubles but not the current?

 

[rl.bhat]

I don't understand. How is the reactance doubles but not the current?

Imax = Vmax/XL = NAB(2ω)/2ωL = NAB/L

 

[Terocamo]

oic the current is independent of rotation speed, thanks.