For a coil of 5 turns, the magnetic flux varies with time as $\phi_{B}(t) = (2t^3 - \frac{5}{2}t^2 + t)$, where $\phi_{B}(t)$ and t are in S.I. units. The induced current at $t=2s$ in the coil, if its total resistance is 150 Ω, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.5 A
Induced emf $(ε)$ is given as -
$ε=-N\frac{d\phi_B}{dt}$
$\left.\frac{d\phi_B}{dt}\right|_{t=2}=\frac{d}{dt}(2t^3+\frac{5}{2}t^2+t)$
$=[6t^2-5t+1]_2$
$=6(1)^2-5(2)+1$
$=24-10+1=15$
$∴ε=-N\frac{d\phi_B}{dt}=5×15=75$
Also, Induced Current (I) is,
$I=\frac{ε}{R}=\frac{75}{150}=0.5A$