Coil P has \(N\) turns, each of radius \(r\). Coil Q has \(2N\) turns, each of radius \(r/2\). Both coils are placed normal to the same uniform magnetic field, which changes at a steady rate. The ratio of the induced emfs \(\varepsilon_P:\varepsilon_Q\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 : 1
For a coil normal to the field, the flux through each turn is \( BA \). When \( B \) changes, the emf in a coil of \( N \) turns is \( \varepsilon=NA\,dB/dt \). Both coils see the same \( dB/dt \), so \( \varepsilon\propto NA\propto Nr^{2} \).
\( \begin{aligned} \frac{\varepsilon_P}{\varepsilon_Q} &= \frac{N_P\,r_P^{2}}{N_Q\,r_Q^{2}} \\ &= \frac{N\,r^{2}}{2N\,(r/2)^{2}} \\ &= \frac{N\,r^{2}}{N\,r^{2}/2} \\ &= 2 \end{aligned} \)
Hence, \( \varepsilon_P:\varepsilon_Q=2:1 \). Coil Q has twice the turns, but each of its turns has only one-fourth the area.
NCERT (2026-27):
Magnetic Flux – “… Magnetic flux through a plane of area A placed in a uniform magnetic field B … can be written as”
\( \Phi_B=\mathbf{B}\cdot\mathbf{A}=BA\cos\theta \)\( \qquad (6.1) \)
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”
\( \varepsilon=-N\displaystyle\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)