The magnetic flux passing through a metal ring varies with time $t$ as : ${\varphi }_B=3(a{t}^3-b{t}^2)T-m^2$ with $a=2.00s^{-3}$ and $b=6.00s^{-2}$. The resistance of the ring $3.0\Omega$. Determine the maximum current induced in the ring during the interval from $t=0$ to $t=2s$.

Given, ${\varphi }_B=3(a{t}^3-b{t}^2)$

$\left|e\right|=\left|\frac{d{\varphi }_B}{dt}\right|=9a{t}^2-6bt$

Induced current, $i=\frac{\left|e\right|}{R}=\frac{9a{t}^2-6bt}{3}=3a{t}^2-2bt$

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

$$\begin{gathered} 6at-2b=0 \\ t=\frac{b}{3a} \end{gathered}$$

i.e. at $t=\frac{b}{2a}$, current is maximum. This maximum current is

$$\begin{gathered} i_{max}=3a{\left(\frac{b}{3a}\right)}^2-2b\left(\frac{\mathrm{b}}{3\mathrm{a}}\right) \\ =\frac{b^2}{3a}-\frac{2b^2}{3a}=-\frac{b^2}{3a} \\ \left|I_{max}\right|=\frac{b^2}{3a} \end{gathered}$$

Substituting the given values of $a$ and $b$, we have

$i_{max}=\frac{{\left(6\right)}^2}{3\left(2\right)}=6A$