A square loop containing an inductor of inductance 4H and a capacitor of 2F, is fixed in the plane of wires (infinitely long) carrying current $I_{1} = (4 t) A$ and $I_{2} = (2 t) A$ as shown in figure. Choose the CORRECT option(s)

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Maximum charge on capacitor is $\frac{4 μ_{0} a}{π} \ln (2)$

Maximum current in the inductor is $\frac{\sqrt{2} μ_{0} a}{π} \ln (2)$

Maximum current in the inductor is $\frac{μ_{0} a}{\sqrt{2} π} \ln (2)$

Maximum charge on capacitor is $\frac{2 μ_{0 a}}{π} \ln (2)$

answer is A, C.

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Detailed Solution

$ε = \frac{μ}{2 π} a \ln 2 \frac{d}{d t} (I_{1} - I_{2}) ε = \frac{μ_{0}}{π} a \ln 2 ∴ Q_{\max} = 2 C \in ∴ Q_{\max} = \frac{4 μ_{0} a}{π} \ln 2 i_{\max} = C \in ω = \frac{2 μ_{0} a}{π} \frac{1}{\sqrt{2 (4)}} i_{\max} = \frac{μ_{0} a}{\sqrt{2} π} \ln 2$