An inductance coil has a reactance of 100 $$\Omega$$. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45$$°$$. The self-inductance of the coil is:

We are told that the coil offers an opposition of $$100\ \Omega$$ to the alternating current and that the current lags behind the applied voltage by $$45^{\circ}$$. In AC circuit language the total opposition is called the impedance and is denoted by $$Z$$, while the opposition due purely to the inductance is called the inductive reactance and is denoted by $$X_L$$. The phase angle $$\phi$$ between voltage and current satisfies the trigonometric relations

$$\tan\phi = \dfrac{X_L}{R},\qquad\; X_L = Z\sin\phi,\qquad\; R = Z\cos\phi.$$

Here the given quantity $$100\ \Omega$$ is the magnitude of the impedance, so we write

$$Z = 100\ \Omega.$$

The phase angle is given as $$\phi = 45^{\circ}$$, and we recall that

$$\sin 45^{\circ} = \dfrac{1}{\sqrt2}.$$

Using the relation $$X_L = Z\sin\phi$$ we substitute the known values:

$$X_L = 100 \times \dfrac{1}{\sqrt2} = \dfrac{100}{\sqrt2}\ \Omega = 70.7\ \Omega$$

(the numerical value $$100/\sqrt2$$ is $$70.7$$ to three significant figures).

Next we connect the inductive reactance to the self-inductance by the standard formula

$$X_L = \omega L,$$

where $$\omega$$ is the angular frequency of the applied signal. Angular frequency is related to ordinary frequency $$f$$ by

$$\omega = 2\pi f.$$

The problem states that $$f = 1000\ \text{Hz}$$, so

$$\omega = 2\pi \times 1000 = 2000\pi\ \text{rad s}^{-1}.$$

Now we solve for the self-inductance $$L$$:

$$L = \dfrac{X_L}{\omega} = \dfrac{70.7}{2000\pi}\ \text{H}.$$

Carrying out the division,

$$L = \dfrac{70.7}{6283}\ \text{H} \approx 1.125 \times 10^{-2}\ \text{H}.$$

Writing this with two significant figures gives

$$L \approx 1.1 \times 10^{-2}\ \text{H}.$$

Hence, the correct answer is Option A.