The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :

Electromagnetic Induction PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q54 with Solution

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Correct answer: C

When a solenoid of inductance $L$ and resistance $R$ is connected to a battery, the current grows as

I(t)=I_0\left(1-e^{-Rt/L}\right)

where the maximum (steady-state) current is

I_0=\frac{V}{R}.

The magnetic energy at time $t$ is

U(t)=\frac{1}{2}LI^2(t).

Its maximum value is

U_{\max}=\frac{1}{2}LI_0^2.

So,

\frac{U(t)}{U_{\max}}=\left(\frac{I(t)}{I_0}\right)^2 =\left(1-e^{-Rt/L}\right)^2.

We are given

U(t)=25\%\text{ of }U_{\max}=\frac{1}{4}U_{\max}.

Hence,

\left(1-e^{-Rt/L}\right)^2=\frac{1}{4}.

Taking the positive root (since current is increasing from $0$ to $I_0$ ),

1-e^{-Rt/L}=\frac{1}{2}.

So,

e^{-Rt/L}=\frac{1}{2}.

Taking natural log,

-\frac{Rt}{L}=\ln\left(\frac{1}{2}\right)=-\ln 2.

Therefore,

t=\frac{L}{R}\ln 2.

\boxed{t=\frac{L}{R}\ln 2}

So the correct option is C.