Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

(N/A) An inductor opposes the flow of current through it by developing a back emf according to Lenz's law. The induced voltage has a polarity that opposes the change in current.
The induced emf is given by $\varepsilon = -L \frac{di}{dt}$. For an alternating current $i = I_0 \sin(\omega t)$,the induced emf is $\varepsilon = -L \frac{d}{dt}(I_0 \sin(\omega t)) = -L I_0 \omega \cos(\omega t)$.
The magnitude of the induced voltage is proportional to the angular frequency $\omega = 2 \pi f$. Since the reactance $X_L$ is defined as the ratio of the peak voltage to the peak current,we have $X_L = \frac{V_0}{I_0} = \omega L = 2 \pi f L$.
As the frequency $f$ increases,the rate of change of current $\frac{di}{dt}$ increases,leading to a higher induced back emf. Consequently,the inductor offers greater opposition (reactance) to the flow of current at higher frequencies.