A coil of 100 turns and area 50 cm2 is held normal to a uniform magnetic field of 0.2 T. In 0.1 s the field is changed to 0.2 T in the opposite direction. The average emf induced in the coil during this time is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\ \mathrm{V}\)
The flux through each turn starts at \( +BA \) and ends at \( -BA \), because the field reverses. So the size of the change in flux per turn is \( 2BA \).
\( N=100 \)
\( A=50\ \mathrm{cm^{2}}=5\times10^{-3}\ \mathrm{m^{2}} \)
\( B=0.2\ \mathrm{T} \)
\( \Delta t=0.1\ \mathrm{s} \)
\( \begin{aligned} \Delta\Phi_B &= BA-(-BA) \\ &= 2\times0.2\times5\times10^{-3} \\ &= 2\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
\( \begin{aligned} \varepsilon &= \frac{N\,\Delta\Phi_B}{\Delta t} \\ &= \frac{100\times2\times10^{-3}}{0.1} \\ &= 2\ \mathrm{V} \end{aligned} \)
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)