Two coils P and Q have the same area. Coil P has 100 turns and a resistance of 2 Ω. Coil Q has 200 turns and a resistance of 6 Ω. The same change of magnetic flux through each turn is produced in both coils in the same time. The ratio of the induced emfs \(\varepsilon_P:\varepsilon_Q\) and the ratio of the induced currents \(I_P:I_Q\) are, respectively,
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1:2\) and \(3:2\)
The induced emf depends on the number of turns and on the rate of change of flux through each turn. It does not depend on the resistance of the coil. The resistance decides only how much current this emf drives.
Each turn of both coils has the same \( d\Phi_B/dt \). So \( \varepsilon=N\,d\Phi_B/dt \) is proportional to \( N \), and \( I=\varepsilon/R \).
\( \begin{aligned} \frac{\varepsilon_P}{\varepsilon_Q} &= \frac{N_P}{N_Q} \\ &= \frac{100}{200} \\ &= \frac{1}{2} \end{aligned} \)
\( \begin{aligned} \frac{I_P}{I_Q} &= \frac{\varepsilon_P/R_P}{\varepsilon_Q/R_Q} \\ &= \frac{\varepsilon_P}{\varepsilon_Q}\times\frac{R_Q}{R_P} \\ &= \frac{1}{2}\times\frac{6}{2} \\ &= \frac{3}{2} \end{aligned} \)
Therefore, the emfs are in the ratio 1 : 2 and the currents in the ratio 3 : 2.
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)