A closely wound circular coil of area 200 cm2 is placed with its plane normal to a uniform magnetic field. The field increases steadily at 0.5 T s−1, and an emf of 0.5 V is induced in the coil. The number of turns in the coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 50
Each turn of a closely wound coil has the same change of flux. So the emf of the coil is \( N \) times the emf of one turn. With the plane normal to the field, the flux through one turn is \( BA \), and the size of the emf is \( \varepsilon=NA\,(dB/dt) \).
\( A=200\ \mathrm{cm^{2}}=0.02\ \mathrm{m^{2}} \)
\( dB/dt=0.5\ \mathrm{T\,s^{-1}} \)
\( \varepsilon=0.5\ \mathrm{V} \)
\( \begin{aligned} N &= \frac{\varepsilon}{A\,(dB/dt)} \\ &= \frac{0.5}{0.02\times0.5} \\ &= 50 \end{aligned} \)
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)