A $50-$turn, $4\mathrm{.}0-$cm$-$diameter coil with R$=0\mathrm{.}50$R$=0\mathrm{.}50$ surrounds a $2\mathrm{.}0-$cm$-$diameter solenoid. The solenoid is $20$ cmcm long and has $200$ turns. The $60$ Hz current through the solenoid is Isol$=(0\mathrm{.}50$A$)$sin$(2$ft$)$Isol$=(0\mathrm{.}50$A$)$sin$(2$ft$).$