A solenoid of 10 Henry inductance and 2 ohm resistance, is connected to a 10 volt battery. In how much time the magnetic energy will be reaches to$\frac{1}{4}th$ of the maximum value?

The net magnetic field is the total of all the loops and is greatest near the centre of the solenoid since that location reduces the average distance between each loop.

Given, L=10H, R=2 ohm and V=10 volt

the maximum current is given by, $i_0=\frac{V}{R}$

$i_0=\frac{10}{2}=5A$

The maximum energy is given by

$E_0=\frac{1}{2}L{i^2}_0$

$E_0=\frac{1}{2}\times 10\times 5\times 5$

$E_0=125J$

The magnetic energy is given by,

$E=\frac{E_0}{4}$

$E=\frac{125}{4}J$

$Also E=\frac{1}{2}L{i}^2$

$\frac{125}{4}=\frac{1}{2}L{i}^2$

$\frac{125}{2\times 10}=i^2$

$i^2=\frac{25}{2}$

$i=\frac{5}{2}=2.5A$

time taken to rise the current from 0A to 2.5A,

$i=i_o[1-e^{-Rt/L}]$

$\Rightarrow 2.5=5[1-e^{-Rt/L}]$

$\Rightarrow \frac{-Rt}{L}={\log }_e(0.5)$

$\Rightarrow e^{-Rt/L}=0.5$

$\Rightarrow \frac{Rt}{L}=0.693$

$\Rightarrow t=6.932$

$\Rightarrow t=3.46$

$\approx 3.5 sec$

Hence the correct answer is option A.