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Step-by-step Solution
Step 1: Express the maximum magnetic energy
The maximum magnetic energy stored in an inductor (solenoid) when the current reaches its steady value $I_0$ is given by
$U_{0} = \tfrac{1}{2} L\, I_{0}^{2}$.
Here, $L$ is the inductance of the solenoid, and $I_0$ is the final (steady) current in the circuit.
Step 2: Relate the given fraction of magnetic energy to the current
The problem states that the magnetic energy has reached 25% (i.e., $\tfrac{1}{4}$) of its maximum value. Let $U$ be the magnetic energy at time $t$. Then,
$U = \tfrac{1}{4}U_0.$
Since $U = \tfrac{1}{2} L\, I^{2}$, we set it equal to $\tfrac{1}{4}\left(\tfrac{1}{2} L\, I_{0}^{2}\right)$:
$\tfrac{1}{2} L\, I^{2} = \tfrac{1}{4}\bigl(\tfrac{1}{2} L\, I_{0}^{2}\bigr).$
Canceling common factors gives $I^{2} = \tfrac{I_0^{2}}{4}$, so $I = \tfrac{I_0}{2}.$
Step 3: Write down the growth of current in an LR circuit
When a constant voltage (battery) is connected across an inductor $L$ and resistor $R$ in series, the current $I$ as a function of time $t$ (starting from $0$ current) is
$I(t) = I_0\bigl(1 - e^{-t/\tau}\bigr) \quad\text{where}\quad \tau = \tfrac{L}{R}\,.
Step 4: Impose the condition for the current to be half of $I_0$
We want $I(t)$ to be $\tfrac{I_0}{2}$, so set
$\tfrac{I_0}{2} = I_0\bigl(1 - e^{-t/\tau}\bigr).
Dividing both sides by $I_0$ yields
$\tfrac{1}{2} = 1 - e^{-t/\tau}.
Hence
$e^{-t/\tau} = \tfrac{1}{2}.
Step 5: Solve for time $t$
Taking the natural logarithm on both sides,
$- \tfrac{t}{\tau} = \ln\Bigl(\tfrac{1}{2}\Bigr) = -\ln(2)
\quad\Longrightarrow\quad
\tfrac{t}{\tau} = \ln(2).
Since $\tau = \tfrac{L}{R}$, we get
$t = \tau\,\ln(2) = \tfrac{L}{R}\,\ln(2).
Final Answer
The time taken for the magnetic energy of the solenoid to reach 25% of its maximum value is
$\displaystyle \tfrac{L}{R}\ln 2$.
