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Step-by-Step Solution
Step 1: Identify the Steady-State Current
When the coil (inductor) of inductance $L=1\,\text{H}$ and resistance
$R=100\,\Omega$ is connected to a $6\,\text{V}$ battery, the final (steady-state)
current $i_{\mathrm{final}}$ is obtained from Ohm’s Law:
$i_{\mathrm{final}} = \frac{V}{R} = \frac{6}{100} = 0.06\,\text{A}.$
This is the maximum current the circuit will reach when sufficient time has elapsed.
Step 2: Time for the Current to Reach Half of Its Steady State Value
The current at any time $t$ in an $RL$ circuit (when the circuit is just switched on) is
given by:
$i(t) = \frac{V}{R}\Bigl(1 - e^{-\frac{R\,t}{L}}\Bigr).$
To find the time $t_{1/2}$ at which the current is half its final value, set
$i(t_{1/2}) = \tfrac{1}{2}\,i_{\mathrm{final}}$:
$\frac{1}{2} \cdot \frac{V}{R}
= \frac{V}{R}\bigl(1 - e^{-\tfrac{R\,t_{1/2}}{L}}\bigr).$
Simplifying,
$\frac{1}{2} = 1 - e^{-\tfrac{R\,t_{1/2}}{L}}
\quad\Longrightarrow\quad
e^{-\tfrac{R\,t_{1/2}}{L}} = \frac{1}{2}.
$
Taking the natural logarithm of both sides,
$-\frac{R\,t_{1/2}}{L} = \ln\bigl(\tfrac{1}{2}\bigr) = -\ln(2)
\quad\Longrightarrow\quad
t_{1/2} = \frac{L}{R}\,\ln(2).
Given $L=1\,\text{H}$, $R=100\,\Omega$, and $\ln(2) \approx 0.693,$ we get:
$t_{1/2} = \frac{1}{100} \times 0.693 = 0.00693\,\text{s} \approx 7\,\text{ms}.$
Step 3: Current at 15 ms
The current in the circuit at time $t$ is again:
$i(t) = \frac{V}{R}\Bigl(1 - e^{-\tfrac{R\,t}{L}}\Bigr).$
At $t=15\,\text{ms} = 0.015\,\text{s},$ the exponent term is:
$\frac{R\,t}{L} = \frac{100 \times 0.015}{1} = 1.5.$
Hence,
$i(15\,\text{ms}) = \frac{6}{100}\Bigl(1 - e^{-1.5}\Bigr).
We are given $e^{-1.5} \approx 0.25.$ Therefore,
$i(15\,\text{ms}) = 0.06\bigl(1 - 0.25\bigr) = 0.06 \times 0.75 = 0.045\,\text{A}.
Step 4: Energy Stored in the Magnetic Field at 15 ms
The energy stored in the inductor at any instant is:
$U = \tfrac{1}{2}L\,i^2.
Substituting $L = 1\,\text{H}$ and $i(15\,\text{ms}) = 0.045\,\text{A}$,
$U = \tfrac{1}{2} \times 1 \times (0.045)^2
= \tfrac{1}{2} \times 0.002025
= 0.0010125\,\text{J}
\approx 1\,\text{mJ}.$
Step 5: Final Answers
(a) Time to reach half the steady current $\approx 7\,\text{ms}.$
(b) Energy stored at $t=15\,\text{ms}$ $\approx 1\,\text{mJ}.$
