\(X _{ L }=\omega L =100 \pi \frac{\sqrt{2}}{\pi}=100 \sqrt{2} \Omega\)
Peak current \(I _{0}=\frac{ E _{0}}{ X _{ L }}=\frac{440}{100 \sqrt{2}}=2.2 \sqrt{2} A\)
\(AC\) ammeter reads RMS value therefore reading will be \(I_{\text {ms }}\)
\(I_{\text {mus }}=\frac{I_{0}}{\sqrt{2}}=2.2\,A\)