A coil of inductance 1 H and resistance 100 $$\Omega$$ is connected to a battery of 6 V. Determine approximately:
(a) The time elapsed before the current acquires half of its steady-state value
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on.
(Given $$\ln 2 = 0.693$$, $$e^{-3/2} = 0.25$$)

We have an LR circuit with inductance $$L = 1$$ H, resistance $$R = 100$$ $$\Omega$$, and battery voltage $$V = 6$$ V.

The time constant of the LR circuit is $$\tau = \frac{L}{R} = \frac{1}{100} = 0.01$$ s $$= 10$$ ms. The steady-state current is $$I_0 = \frac{V}{R} = \frac{6}{100} = 0.06$$ A.

(a) Time to reach half the steady-state current: The current in an LR circuit grows as $$I = I_0(1 - e^{-t/\tau})$$. Setting $$I = \frac{I_0}{2}$$: $$\frac{1}{2} = 1 - e^{-t/\tau}$$, so $$e^{-t/\tau} = \frac{1}{2}$$, giving $$t = \tau\ln 2 = 0.01 \times 0.693 = 0.00693$$ s $$\approx 7$$ ms.

(b) Energy stored at $$t = 15$$ ms: At $$t = 15$$ ms $$= 0.015$$ s, we have $$\frac{t}{\tau} = \frac{0.015}{0.01} = 1.5$$. The current at this instant is $$I = I_0(1 - e^{-1.5}) = 0.06(1 - e^{-3/2})$$. Using the given value $$e^{-3/2} = 0.25$$: $$I = 0.06(1 - 0.25) = 0.06 \times 0.75 = 0.045$$ A.

The energy stored in the inductor is $$U = \frac{1}{2}LI^2 = \frac{1}{2}(1)(0.045)^2 = \frac{1}{2}(0.002025) \approx 0.001$$ J $$= 1$$ mJ.

Hence, the correct answer is Option C.