An electrical circuit is shown in the figure. It has two conducting sphere A and B of radii 3a and 6a respectively. When the inductor of inductance L is in its steady state the switch is shifted from position-1 to position-2. Now choose the correct option(s)

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The circuit contains only electrical energy at time $t = \sqrt{2 π^{3} ε_{0} a L}$

The circuit contains only magnetic energy at time $t = \sqrt{8 π^{3} ε_{0} a L}$

Maximum charge on the sphere A is $\frac{V}{R} \sqrt{8 π ε_{0} a L}$

Maximum charge on the sphere B is $\frac{V}{R} \sqrt{4 π ε_{0} a L}$

answer is A, C, D.

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Detailed Solution

$i_{0} = \frac{V}{R}, 3 C = 4 π ε_{0} a \times 3 \Rightarrow C = 4 π ε_{0} a, \frac{d I}{d t} < 0, \frac{q}{3 C} + \frac{q}{6 C} - L | \frac{d I}{d t} | = 0$

$\Rightarrow L \frac{d^{2} q}{d t^{2}} + \frac{q}{2 C} = 0 \Rightarrow \frac{d^{2} q}{d t^{2}} + \frac{q}{2 L C} = 0 \Rightarrow q = q_{0} \sin ω t, I = q_{0} ω \cos ω t, ω = \sqrt{\frac{1}{2 L C}}$

At t = 0,

$I = I_{0} \Rightarrow q_{0} = \frac{I_{0}}{ω} = I_{0} \sqrt{2 L C}, q_{0} = \frac{V}{R} \sqrt{8 π ε_{0} a L}, T = \frac{2 π}{ω} = 2 π \sqrt{8 π ε_{0} a L}, \frac{T}{4} = \sqrt{2 π^{3} ε_{0} a L}$