The magnetic flux through each turn of a 50-turn coil changes with time as \(\Phi=2\times10^{-3}\sin(100t)\ \mathrm{Wb}\), where \(t\) is in seconds. The maximum emf induced in the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 10 V
The emf in a coil of \( N \) turns is \( -N\,d\Phi/dt \), where \( \Phi \) is the flux through one turn. The rate of change of \( \sin(100t) \) is \( 100\cos(100t) \).
\( \begin{aligned} \varepsilon &= -N\frac{d\Phi}{dt} \\ &= -50\times2\times10^{-3}\times100\cos(100t) \\ &= -10\cos(100t)\ \mathrm{V} \end{aligned} \)
The emf is largest when \( \cos(100t)=\pm1 \).
\( \varepsilon_{max}=10\ \mathrm{V} \)
NCERT (2026-27):
Faraday’s Law of Induction – “In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, the expression for the total induced emf is given by”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)