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(A) According to Faraday's law of electromagnetic induction, the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.Giv

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$4$

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(A) According to Faraday's law of electromagnetic induction, the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.Given $\phi = (5t^2 + 4t + 2) 	imes 10^{-3} 	ext{ Wb}$.Differentiating with respect to $t$:$\frac{d\phi}{dt} = (10t + 4) 	imes 10^{-3} 	ext{ Wb/s}$.At $t = 6 	ext{ s}$, the magnitude of induced $EMF$ is:$|\varepsilon| = |10(6) + 4| 	imes 10^{-3} = 64 	imes 10^{-3} 	ext{ V}$.The induced current $I$ is given by $I = \frac{|\varepsilon|}{R}$.Given $R = 16 \Omega$, we have:$I = \frac{64 	imes 10^{-3}}{16} = 4 	imes 10^{-3} 	ext{ A} = 4 	ext{ mA}$.Thus, the correct option is $A$.

$A$ coil of resistance $16 \Omega$ is placed with its plane perpendicular to a uniform magnetic field whose flux ($\phi$ in $10^{-3} \text{ Wb}$) changes with time ($t$ in seconds) as $\phi = 5t^2 + 4t + 2$. The induced current at time $t = 6 \text{ s}$ is: (in $\text{ mA}$)

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