The magnetic flux through a coil of resistance 6 Ω, perpendicular to its plane is varying according to the relation
$\phi= (6t^3+5t^2+4t-6) Wb$
What will be the induced current through the coil at $t = 2 s$?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16 A
Given:
Coil resistance, $R = 6\ \Omega$
Magnetic flux: $\phi(t) = (6t^3 + 5t^2 + 4t - 6)\ \text{Wb}$
Induced emf in the coil:
$\mathcal{E} = -\frac{d\phi}{dt}$
$\frac{d\phi}{dt} = \frac{d}{dt} (6t^3 + 5t^2 + 4t - 6) = 18t^2 + 10t + 4$
So, $\mathcal{E} = -(18t^2 + 10t + 4)\ \text{V}$
Current through the coil: $I = \frac{|\mathcal{E}|}{R} = \frac{18t^2 + 10t + 4}{6}$
At $t = 2\ \text{s}$:
$I = \frac{18(2)^2 + 10(2) + 4}{6} = \frac{72 + 20 + 4}{6} = \frac{96}{6} = 16\ \text{A}$
∴ Induced current through the coil at $t = 2$ s is $16\ \text{A}$