A uniform magnetic field passes normally through the plane of a 50-turn coil. The field is reduced steadily to zero in 0.2 s. An average emf of 0.5 V is induced in the coil during this time. The magnetic flux through each turn of the coil at the start was
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\times10^{-3}\ \mathrm{Wb}\)
The field falls steadily to zero. So the flux through each turn falls from its starting value \( \Phi \) to zero at a steady rate. The emf of the coil is the number of turns times this rate.
\( N=50 \)
\( \Delta t=0.2\ \mathrm{s} \)
\( \varepsilon=0.5\ \mathrm{V} \)
\( \begin{aligned} \varepsilon &= \frac{N\Phi}{\Delta t} \\ \Phi &= \frac{\varepsilon\,\Delta t}{N} \\ &= \frac{0.5\times0.2}{50} \\ &= 2\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)