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 \( t = 2\,\mathrm{s} \) 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
By Faraday's law the induced emf in a coil of N turns is \( \varepsilon = -N\,d\phi_B/dt \), and the current follows from \( I = \varepsilon/R \).
\( \displaystyle \frac{d\phi_B}{dt} = 6t^{2} - 5t + 1 \)
\( \left.\frac{d\phi_B}{dt}\right|_{t=2\,\mathrm{s}} = 6(4) - 5(2) + 1 = 24 - 10 + 1 = 15\,\mathrm{Wb\,s^{-1}} \)
\( |\varepsilon| = N\left|\frac{d\phi_B}{dt}\right| = 5\times15 = 75\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{75\,\mathrm{V}}{150\,\Omega} = 0.5\,\mathrm{A} \)