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(B) When an iron rod is inserted into the inductor,its self-inductance $L$ increases because the permeability of the core increases.The inductive reac

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decreases

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(B) When an iron rod is inserted into the inductor,its self-inductance $L$ increases because the permeability of the core increases.The inductive reactance is given by $X_L = \omega L$. As $L$ increases,$X_L$ also increases.The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$. Since $X_L$ increases,the total impedance $Z$ of the circuit increases.The current in the circuit is given by $I = \frac{V}{Z}$. As $Z$ increases,the current $I$ flowing through the circuit decreases.The power dissipated in the bulb is given by $P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases,and therefore,the glow of the light bulb decreases.

$A$ light bulb and an inductor coil are connected to an $AC$ source through a key as shown in the figure below. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the light bulb

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