Consider an LC circuit, with inductance L = 0 . 1   H and capacitance C = 10 - 3   F , kept on a…

Emf induced in the circuit is

$\left|E\right|=\left|\frac{d\varphi }{dt}\right|=\frac{d}{dt}\left[\left(B_0+\beta t\right)A\right]$

$=\beta \times A$

$=0.04 \mathrm{V}$

So the circuit can be rearranged as

Using Kirchhoff's law we can write

$E=L\frac{di}{dt}+\frac{q}{C}$

$L\frac{di}{dt}=E-\frac{q}{C}$

For maximum current, $\frac{di}{dt}=0$, 

Or $q=CE$

Using work energy theorem,

$qE=\frac{1}{2}L{i}^2+\frac{q^2}{2C}$

At $i_{\max }$, $CEE=\frac{1}{2}L{i}_{max}^2+\frac{E^2{C}^2}{2C}$

Or, $i_{\max }=\sqrt{\frac{C}{L}}E$

So, $i_{\max }=\sqrt{\frac{{10}^{-3}\times 0.04}{0.1\times {10}^{-3}}}$$=4 \mathrm{mA}$